What does the "capacitor into resistance" symbol mean?
It's not one symbol.
It's just a capacitor connected to the wiper terminal of a potentiometer.
Just to explain the potentiometer, since you have your answer now... if you look at one physically, you'll see a carbon track, from the left terminal to the right one. That's a resistance. The middle terminal connects to the wiper, the copper slider that contacts the track. Turning the shaft moves the wiper, varying the resistor.
You seemed to understand that already. In some cases you'd use just 2 terminals as a variable resistor. But it's often more useful to use all three. If you connect, say, 5V to one side, gnd to the other, then the wiper will give a voltage between 5V and gnd, variable.
If you connect one end to gnd, and the other to a signal, the wiper will give a voltage partway between the signal and gnd. As in a volume control.
Often electrical circuits are controlled by a particular voltage somewhere that you want to control. You need the 3 terminals of a potentiometer, aka "pot", to do that. A simple variable resistance would, by itself, just limit the current coming through. That's not always useful. In most circuits, a pot is used with all 3 terminals.
As others say - it's a capacitor on a pot wiper.
Q1 is an inverter with a gain without the mystery pot of about 38 x (V+ - 1) [for reasons related to transistor physics] ~= 300 if V+ = 9V, but with Vc_Q1 amost at ground.
Q2 emitter provides a buffered inverted input signal with is fed back via the left hand 100k pot to stabilise gain. The position of the pot wiper alters the frequency response of the RC feedback network in a probably undesigned and 'interesting' manner.
Wiper far left = a sub 1 Hz low pass filter for feedback PLUS a large cap on the input. so signal probably low. Pot wiper far right - the Q2 emitter follower drives the cap but it again probably clamps the voltage enough to prevent feedback so you get high gain overall plus high gain from Q2 so massive output clipping.
As you slide pot right to left you probably get increasing modification of signal, reduction in overall gain but change in frequency response.
The circuit appears to require enough signal to drive Q1 into conduction, so on very low signals it probably produces no output.
I started to say what that change would be but decided "it's complex" :-). It would sound very bad (or very good with the right ears on).
As a 'bonus' the circuit acts overall as a frequency response modified Schmitt trigger. I won't even start to try to suggest what happens with pot variatioj - but simulation would be interesting.