... I was thinking of a logarithmic 10kΩ pot ...
The gain of IC2 & IC4 are (R9+R10)/R10 and (R22 + R23)/R23 respectively.
IF gain with R9 = 1k2 is "about right" then 10k may be too large - ie you will get about 8x the gain using a 10k pot. A 2k2 or 4k7 may be more appropriate. Linear would be OK enough. Log would better match ear response but either is OK.
Extra: You could either replace R9 with a pot with wiper tied to one end OR could replace both R9 and R10 with a pot, one end of track to IC2-6, other end of track to ground and wiper to IC2-2. The disadvantage of this is that when wiper goes to ground Gain is "infinite" (or open loop.)You could avoid this by placing a resistor between ground and the lower end of the track - but may as well then just use the pot in place of R9.
The arrow "-VA" at IC1 (pin 4) should connect to the "-VA" point above C2, right?
I think you meant to write "above C7" - in which case, yes. This is op amp negative power supply , and R7, C13, C9 form a ripple filter. The resistor should almost be unnecessary as long as the capacitors are used and are close to IC1, IC2 physically - but should do no harm.