Well you are right... Capacitor charges up and then stops. From that moment
it is like open circuit and it does not affect the potential on either side.
When mic picks some noise or sound voltage accross mic will change.
Depending on direction of change, C1 will be either further charged or
discharged until it's in ballance again.
This charging/discharging current is proportinal to signal comming from
microphone. This charging current will temporary create imbalance on the
side connected to transistor. Because audio signal is too fast for capacitor
to fully charge or discharge, the changing audio signal comming from mic
is transferred through capacitor. Note that only CHANGE is transferred.
Change in signal is considered the "AC component" of the signal.
"Full signal" on the microphone side is combination of DC signal
(ca 2-3V depending on model of the microphone) and the changing signal
(few milivolts comming from microphone).
Full signal on the base of the Q1 transistor is combination of the
DC voltage (ca 0.7V) and AC component which is "comming through" C1
(few milivolts).
C1 is sized to be large enough so the AC component of the signal can
pass.
Impedance of capacitor is calculated like Xc=1/(2*pi*f*C) where
pi = 3.14152
f = frequency of the signal
C = capacitor value
If the frequency increases, impedance is lower. In other words, the
higher the frequency of the signal, the smaller capacitor you can use
to pass the signal through.
C1 is used to transfer audio frequencies. C4 and C5 are also used in
a similar way - to transfer some of the AC signals to ground
(high frequencies such as audio noise and interference).
That's why C5 is much smaller than C1. If you increase values of
C4 and C5, the audio signal will be shorted to ground as well and you
will hardly hear anything from your radio.
Q2 works as radio frequency oscillator (much higher frequency)
so the capacitors arount it are much smaller.
As for D1 and C3, I'd just get rid of them and put jumper.