How does electric current flow through a potentiometer?
In the picture, if I move the wiper (B) down - further from A - will
  the resistance increase (and hence voltage drop increase) between A
  and B meaning a lower voltage will be supplied to the power amplifier.


Yes.

Perhaps it would help to consider the input impedance of the power amplifier to be very large (much greater than the potentiometer). As such, any small amount of current will develop a sizable voltage (labeled above as Vin).

To control this we could place a potentiometer (consider points A and B only and disconnect point C from ground) in series between the preamplifier and power amplifier.  However, since the resistance of this potentiometer (50Kohms) is likely much smaller than the input impedance of the power amplifier, the effect of turning the potentiometer is negligible.  

So, instead, lets setup a path for the current from the pre amplifier with a lower resistance.  We can use the potentiometer (now consider points A and C) for this lower resistance path.

Now, if the wiper of the potentiometer was moved all the way to the "A" side of the potentiometer, Vin would be about (ignoring the impedance of the power amplifier's input) the current from the preamplifier times 50Kohm.  This is our maximum volume setting.  Let's call that Vmax.  If the wiper were moved half way between the "A" and "C" sides of the potentiometer, Vin would be cut in half.  That is:

Vin = Vmax * (25Kohms / 50Kohms) = Vmax * 1/2

And if we move the wiper all the way to the "C" side, Vin would be zero:

Vin = Vmax * (0ohms / 50Kohms) = Vmax * 0

This would be the minimum volume setting.  You shouldn't hear a thing out of the power amplifier.