You wrote:
On the way to the load resistance, we maximize voltage transferred but
when in front of the load resistance, we maximize power transferred by
matching the impedances
and this seems to imply that you believe that the maximum power on the load will be transferred when the impedances are matched. Well, this is not necessarily the case. In fact it is in general not true.
Let me take a level of complexity out of the equation by considering resistances and not impedances, to make my case.
Impedance matching does NOT imply the load gets the highest power
Now, suppose you are given a 1 ohm load and you want to power it with a 10 volts battery, hopefully getting it to dissipate a power close to V^2/R = 100W. You go to the mall and find three 10V batteries each with a different internal resistance (it's a strange shop, not only the sell batteries with nonstandard voltages, but they also state their internal resistance...).
The first battery has an internal resistance of 0.1 ohms, the second 1 ohm, the third a whooping 10 ohms (for some reason they are in discount).
Which battery will maximize the power transfer to your 1 ohn load?
Certainly not the one with the humongous internal resistance. But not even the one with the 'matched' resistance of 1 ohm. In fact, unsurprisingly, it's the 0.1 ohm battery that will deliver the highest power to RL.
V = 10 V, Rs = 0.1 ohm, RL = 1 ohm
PRL = 82.6 W
V = 10 V, Rs = 1 ohm, RL = 1 ohm
PRL = 25 W
V = 10 V, Rs = 10 ohm, RL = 1 ohm
PRL = 0.82 W
So, what is this maximum power transfer theorem about then?
Well, consider the battery alone, what is the maximum power it can develop (not deliver, develop)? Short circuit it and you will see all its power going into heat, dissipated by its internal resistance (representing complex internal processes we are not interested in). For the three batteries above, the short circuit powers are
- Rs = 0.1 ohm, PCC = 1000 W
- Rs = 1 ohm, PCC = 100 W
- Rs = 10 ohm, PCC = 10 W
the problem is that in the best of cases (yes, when there is matching) only a fourth of this power can be delivered to the load. So, in the first case you can get at most 250W on a 0.1 ohm load, in the second 25W on a 1 ohm load, and in the third 2.5W on a 10 ohm load.
If you choose to buy a battery that matches your load, i.e. the one with an internal resistance of 1 ohm, well congratulations, you are satisfying the maximum power transfer theorem but instead of the ideally maximum 100W on your load you are getting only 25W. Yes, 25W is the maximum you can get for that kind of battery, but this is a meager satisfaction since your load is underpowered.
A few formulas
To see what is going on graphically, let's consider the expressions for the power dissipated on the load and on the internal resistance in a circuit constituted by a voltage generator with internal resistance Rs that creates a voltage divider with the load RL. By choosing rms values we can pretend we are still in a DC condition and the formulae are:

If we focus on the power dissipated by the load we see that it has two different functional forms, depending on which parameter we consider it function of. If we see it as a function of RL it has a bell shaped form with a peak for RL = Rs (yes, it's actually a bell if you plot it on log scale),

but if we see it a function of Rs, it is a monotonically decreasing function (for Rs>=0) that has a finite maximum in Rs=0.

So, the choice RL = Rs maximize power on RL when Rs is given and you are seeing the power as a function of RL, but if you are given RL and you can choose Rs, then the value that maximize power on RL is Rs=0, and NOT Rs=RL.
Incidentally, the power dissipated by Rs has a dual functional form and when you see PRS as a function of RL you will see that the choice Rs = RL is what maximize the power lost by Rs.
Here are the plots (sorry, here Rs is called Rout) for the powers dissipated by RL (in blue) and Rs (in red) as functions of RL

and here are the same powers seen as functions of Rs (called Rout)

Finally, here is the power dissipated by RL as a function of both RL and Rs (it is still called Rout, and the numerical values used for this graph are different from the two-dimensional ones - just for aesthetics)

Impedance matching runs along the RL=Rs line in the horizontal plane but it's not necessarily the choice that will deliver the highest power on RL.
So, why doing impedance matching in the first place?
Well, from the point of view of power, you want to match the impedances when you want to extract all the juice you can from the source.
Getting back at the battery example: the best of the three choices that gives the higher power to the load is not the matched impedance one, but the one with Rs=0.1 ohm, which delivers 82.6 W on RL. Even if this is the closest to the nominal 100W you would expect applying 10V on a 1 ohm load, this is just a tiny fraction of the maximum deliverable power of 250W for a battery of that kind. But even if I'm just extracting a fraction of what I could, if my goal is to get the nominal power to my load (irregardless of how efficiently I am squeezing juice out of it) that's the best choice.
If I wanted to suck all I can from that battery, I'd have to choose a different load, one that matches its internal impedance, i.e. a 0.1 ohm load; in that case I would be able to suck all the deliverable 250W.
- So, if the source was a solar cell, I'd do my best to get all the
juice out of it by design the first stage of my solar battery charger
in such a way as to match its impedance - not to let it just sit
there in the sun.
- In the battery example above, I'd go for impedance
matching if I were to do a youtube video on melting metals with a car
battery: too low a load and I'll just boil the battery, too high the
load and nothing spectacular happens (even if the load is drawing
almost all of its nominal power), match RL = Rs,and maximum danger
and fireworks ensue.
If you are working in AC and have to juggle impedances, matching has the added bonus of compensating the reactances of Zs and ZL.
If you are into RF, matching has a completely new reason to exists since it eliminates the reflections that happen when there is an impedance mismatch, thus reducing or removing signal integrity problems, and inefficient power transmission.
And sometimes matching is required by sheer compliance to standards. Imagine a world where function generators and electronic instrumentation had random impedances. Even if you did not reach frequencies high enough where reflections could be a problem, you would still have the divider problem: your scope has 13 ohm input resistance, your generator has 74 ohm output impedance... what voltage will you see? And when you use another function generator with 123 ohms output impedance? Madness. Let's get a reasonable standard value - or a limited set of such values - and everything is easier.
In amplifiers, well, usually you have to know what you want to amplify. If it's voltage, you want the highest input impedance possible while if you want to amplify current you will look for the lowest input impedance. Yes, power drawn will be negligible, but you can count on the last stage to deliver the right amount of power to your load. That's what the last buffer is for: you amplify voltage along the chain, then 'add' current.
Sometimes you add in a stage whose sole purpose is to translate a low impedance into a high impedance (or vice versa). The power supplied to the stage will provide the extra current or voltage required to avoid losing dBs.
Of course, in RF amplifiers you might want to impedance-match each stage, but the threshold frequency above which this make sense depends on the scale of integration (as described in another answer).