We don't have to understand why the transistor exhibits current gain in order to understand amplification. Those topics can be understood separately. To understand the amplifier, we can regard the transistor as a "black box" which has certain behaviors which can be described with math functions and plotted as graphs.
If you approach an amplifier circuit from the perspective of majority carriers and doped regions in the transistor, it will be unnecessarily confusing.
Also, to understand the transistor circuit, we should delay thinking about AC and first concentrate on DC. There are DC amplifiers, which should be understood first! The AC amplifier essentially works by superimposing back-and-forth disturbances on the quiescent point of a DC amplifier. (But the AC amplifier is optimized toward AC operation, which is necessary because AC signals "see" different impedances in a circuit from DC. The AC amplifier has a different load line and gain from the DC amplifier which it is based on. These are complications which are easier to understand once you have a grasp on the DC amplifier.)
Voltage amplification takes place when the output is taken from a collector-side load resistance. Changes in the base voltage cause corresponding changes in the much larger current through the collector. The fluctuations in the collector current are multiplied by the load resistance on the collector side to create a fluctuating voltage (V = IR). If the amplifier is biased right, that voltage is larger than the input voltage.
If the output is taken from the emitter resistor instead, then there is no voltage amplification: the amplifier is a voltage follower. But in that case, still the transistor provides a boost to the signal. Even though the output voltage tracks the input voltage, the transistor can supply more current, so it can drive heavier loads (lower impedances). It is is a power amplifier.
Why does the NPN transistor exhibit current gain? The simple explanation "for dummies" is that the behavior occurs because the base region is very thin. Electrons are driven by the forward bias of the base-emitter diode to flow from the emitter terminal to the base terminal, which requires them to traverse the emitter and then the base. Because the base is lightly doped, the electrons have a hard time reaching the base terminal. They have to "go slow". And because the base is thin, the electrons are forced to travel across the base close to the collector. It's as if the electrons are asked to walk on a rope. So, most of them "slip and fall" into the collector instead before they are able to escape to the base terminal. Hence, the emitter has to supply many extra electrons in order to sustain the base current, and most of the emitted electrons go into the collector.