Electronic – Hi-Fi Stereo Amplifier gain regulation with potentiometer

... 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.

Each of the 3 power rails (3.3V, 5V, +12V) are connected to a fuse, then to a binding post, while the negative binding posts go to ground.

This sounds like the problem. Keep in mind, voltages are differences in potential between two points. There's nothing special about ground, it's just an arbitrary point which we pick. It's 0V because the difference of something with itself is 0.

You can call anything you want "ground". That's why this circuit works. If you call the -12V output as "ground", then everything else is 12V higher. That's including what was previously called "ground": now it's 12V, because it's 12V more than what you are now calling ground.

Now consider what you've done:

^{simulate this circuit – Schematic created using CircuitLab}

The "ground" of the power supply connects all of the voltages supplied together (connections labeled B and C). The output voltages are relative to this. Notice how the -12V (V4) makes a negative voltage because it's positive side is attached to "ground".

Then, you attached the negative binding posts of all the supplies together. Largely this is redundant: you are duplicating connections B and C. But you are also adding connection A.

See the problem? You've shorted out V4. A wire has ideally zero resistance. By Ohm's law, the current that will flow is:

$$ \frac{12V}{0\Omega} = $$

In reality, the wires used to make this connection actually have some very small resistance, and you get a whole ton of current. This far exceeded the current the voltage regulator can handle and the smoke got out.