I dedicate this story to @jonk - the amazing professional, enthusiast and Great Man with the soul of a child, with the hope that he will come back to us!
Understanding through concepts
No matter how hard we try to explain something in detail, for it to be truly understood, we need to show the basic ideas... the concepts on which it is built.
In electrical circuits, such powerful concepts are resistor voltage-to-current converter, resistor current-to-voltage converter and their derivatives - variable resistor (rheostat) and voltage divider (potentiometer). They have been well known since the 19th century; we use them for their purpose now... and they will be used in the future because they are eternal and immortal.
By these concepts we can indirectly explain the more abstract transistor circuits. Indeed, electrical resistances are linear and transistor resistances are non-linear, but this does not particularly matter for the intuitive understanding of the most basic principles on which transistor amplifier stages are built. Furthermore, non-linear resistances can be represented as changing (static) resistances and thus a connection between electrical and transistor circuits can be made.
That is why, in order to truly answer the OP's fundamental question, I have examined, step by step, the evolution of the elementary transistor stage in parallel with its electrical (conceptual) implementation. I have done it by the help of CircuitLab simulations but I have also created a similar story with real experiments.
Building transistor amplifier
STEP 1: Controlling current by a variable resistor
The amplification is a kind of regulation. The simplest form of such an operation is the current control. We can do it, according to Ohm's law, by changing either the voltage or the resistance or both. In amplifiers we use the second way.
Resistor current "amplifier". So, to regulate the current in the simplest Ohm circuit, we insert a variable resistor Rin and start changing its resistance. Thus, the resistance is the input quantity of our "amplifier" and the current Iout is the output quantity. Let's set an initial (quiescent) current value of 5 mA. We can do it in two ways - by calculation from Ohm's law or by adjustment (changing Rin in parameters window while looking at the ammeter).

simulate this circuit – Schematic created using CircuitLab
It is interesting to see the transfer curve Iout = f(Rin) of our "amplifier". As you can see, it is non-linear because the input quantity (the resistance) is in the denominator but if we change Rin within small limits, the dependence will be close to linear.

It would also be better to start with the largest resistance value (lowest current value), but the CircuitLab DC sweep simulation does not allow this.
Transistor current amplifier. Transistors can be functionally thought of as a type of "resistors" which, like ordinary resistors, dissipate heat and power, and limit the current. So let's replace the "manually-controlled resistor" Rin with such an "electrically-controlled resistor" - the transistor Q. Now the voltage Vin applied to the base-emitter junction is the input quantity of our "true amplifier" and the current Iout (Ic) is the output quantity.

simulate this circuit
A little more dexterity is required to set the quiescent current of 5 mA since at some point (about 0.6 V), the base-emitter junction begins behaving as a voltage-stabilizing non-linear resistor - it decreases its resistance when we increase the voltage and v.v. As they say, it has "low differential resistance"; I would say it behaves as a "dynamic resistor". Practically, we can adjust Vin as above - opening the Vin parameters window and changing Vin while looking at the ammeter.
The transfer curve Iout = f(Vin) of our amplifier is non-linear again, now because of the base-emitter junction properties. But we can apply the same linearization trick as above - if changing Vin within small limits ("wiggling" around the quiescent input voltage aka "bias voltage") the dependence will be close to linear (as they would say, "Vin is a small-signal quantity").

STEP 2: Controlling voltage by a voltage divider
The amplifier we invented above produces output current... but we need a voltage output. Ohm's law gave us the idea above to use a resistor as a voltage-to-current converter (I = V/R). Now it can give us another idea - to use a resistor as the inverse current-to-voltage converter (V = I.R).
Resistor voltage "amplifier". So insert a constant resistor Rc in the place of the ammeter (or, a clever trick, increase the ammeter resistance in its parameters window). 1 k is a very convenient value because it eliminates the need for calculations (another clever trick).

simulate this circuit
Now you can set the quiscent voltage Vout = 5 V by changing Rin in its parameters window while looking at the voltmeter. The result is interesting - Rin = Rc, i.e. the two resistors form a voltage divider with a ratio of 0.5. So our "resistor amplifier" is actually a voltage divider with controlled ratio.
The transfer curve Vout = f(Rin) of the "resistor voltage amplifier" is the same as of the "resistor current amplifier" above because of the clever 1 k trick - [V] = [mA].[kohm].

Transistor voltage amplifier... Now let's improve the transistor current amplifier above by converting the output current into voltage.
... by floating output... As in the resistor amplifier, we connect a 1 k "collector" resistor Rc and take the voltage drop across it as an output voltage Vout. As above, a voltage divider is formed where the top resistor is the same but the bottom resistor is a transistor. If you compare the two arrangements - this and the resistor "amplifier" above, you will see that they are functionally identical. The transistor behaves as a 1 k "resistor" with the same voltage as the voltage across Rin above and the same current through it (still there is some difference and to find out what it is, change Rc in both configurations).

simulate this circuit
When plotting the transfer characteristic Vout = f(Vin), we encounter a difficulty because the DC sweep simulation cannot directly plot a "floating" voltage such as Vout. Therefore, we ask it to plot the result of the expression Vout = Vcc - Vc.

... by grounded output. Transistor amplifiers and electronic circuits in general are cascaded devices where the input of the next device is connected to the output of the previous one. As a rule, their inputs and outputs are "single ended" (grounded). However, the output Vout of our amplifier is referenced to Vcc. What do we do?
They asked themselves this question many years ago and solved it elegantly. It is much easier for us now... We can even get to this idea via the CircuitLab trick above (Vout = Vcc - Vc). Let's formulate it:
Instead of the referenced to Vcc voltage VRc, we use its referenced to ground complement Vcc - VRc.

simulate this circuit
Accordingly, the graph of the transfer curve is complementary to the graph of the stage with referenced to Vcc output voltage. It simply represents the Q collector voltage Vc.

This famous configuration has two features:
Vout is not the same in voltage value but its change value is the same. An AC amplifier, for example, amplifies only the change in voltage.
The direction of change is reversed - when VRc increases in value, Vout decreases. That is why this amplifier stage is called "inverting". Note that Vout has the same polarity as Vin but its direction of change is opposite.
STEP 3: Applying emitter voltage
The transistor amplifier we "invented" above has a grounded emitter; so its input (transistor base) is single-ended (referenced to ground). However, the transistor has the unique ability to have its emitter used as a second input. Thus, something like an "amplifier with a differential input" (the base-emitter junction) is obtained. I have known this for a long time, but only now have I realized why it is possible. I will say it briefly here and we will see it in the experiments below.
The reason the emitter can be used as an input is that the transistor output (its collector-emitter part) behaves as a current source that creates the voltage drop VRc across the collector resistor and, accordingly, its complement Vout referenced to ground. That is why, this current and accordingly, the output voltage, does not depend on the emitter voltage.
Driving the transistor through the base (common-emitter stage)
For starters, let's "lift" the Q's emitter with some constant voltage Vref. Since we chose a quiescent voltage Vc = 5 V, we need to leave enough "room" for the voltage Vce to change (so that the transistor does not saturate). OK, let's set 1 V and raise the input voltage Vin = 0.701 V of the grounded-emitter amplifier above by that much. Thus the difference Vbe between the two voltages remains the same, the same collector current Ic = 5 mA flows through the resistor Rc and accordingly, the same output voltage Vout = 5 V appears.

simulate this circuit
As we expected, the graph of the transfer curve is almost the same as the graph of the stage with grounded emitter. The only difference is that the transistor saturates earlier - when its collector voltage becomes 1 V (because its emitter is lifted by 1 V). Our conclusion is that there is no difference between the two configurations; only the output voltage range is reduced in the second one.

Driving the transistor through the emitter (common-base stage)
But agree that with the same success we can drive the transistor from the emitter. Why not? We just start varying the emitter voltage Vin as an input while keeping the base voltage Vref constant; their difference is the same as above.

simulate this circuit
Looking at the graph, we can see that:
Regarding the gain, the two configurations are the same because Vbe is the same.
The output voltage changes in the same direction as the input voltage, i.e. the circuit is non-inverting; the reason is that in the input loop Vref reverses the sign of Vbe.
The output voltage cannot go lower than the input voltage because the transistor saturates.
The output (collector) current passes through the input source trying to change its voltage in the opposite direction. So a negative feedback will appear if the source has some internal resistance.
