Q:

Is my understanding of how this amplifier circuit works correct?

Hello.  This is the amp that I'm trying to wrap my head around.
This is how I understand how it works and I'd love your input if what I say makes sense:
The input voltage is a sine-wave of increasing and decreasing voltage.  Let's say that the highest voltage is 4 and the lowest is 3.  When this input hits the initial capacitor, this voltage drops down to 1 to 0.  Basically, the power is either on or off.
When current is flowing, this causes the NPN transistor to allow power to flow through it and the electrons from the ground to link up with the electrons flowing through R2 and R4 to create an amplification.  However, when the voltage drops to 0, then the NPN stops the transfer of power entirely and the output is whatever the electrons coming across from the input signal.
Does this make sense or am I totally off-base?

A:

Sorry, but that's way off. 
Start with R1 and R2, and no input signal. Then R1/R2 form a voltage divider which sets the operating point for the transistor. It is assumed that the circuit is set up so that the base current is much less (< 1/10) the unloaded current through R1/R2.
Now put a sine wave of (let's say) 3.9 to 4.1 volts on the input. The capacitor will indeed get rid of the 4 volt offset, but it will reference the AC to the operating point voltage. So let's say that +V is 6 volts, and R1 = R2. The operating point is 3 volts, and the capacitor will shift the input to 2.9 to 3.1 volts at the base of the transistor.
This voltage will be a sine wave, and this will cause the output of the transistor to be (more or less) a sine wave as well.

