Hm, I did a simulation too, just up through the rightmost diode.
Except I assumed the input signal was synth level so used a 10 Vpp sine, in which case the first op amp output is almost always ±12 V, and what comes off the pot wiper is a near square wave. (Note the pot is configured like a voltage divider, but there’s current through the wiper when the voltage is above ~700 mV.)
I think what’s happening is:
When the pot wiper voltage is negative the leftmost diode has only a small leakage current. That current has to come through the top left diode, so the second op amp output must be a diode drop above -(the wiper voltage), and the output is a diode drop below that.
When the pot wiper voltage is positive so is the inverting input on the second op amp; it has no way to change that, so the op amp output is -12 V. Then the output is three diode drops below the wiper voltage.
I don’t really see the point of the rightmost diode other than to reduce the output by a diode drop; it’s always in conduction (for ±12 V out of the first op amp).
Output for different pot settings:
Why they’re going through all that just to drive a vactrol I don’t know, it’ll presumably just smooth this out into an envelope, so it’s kind of an envelope follower but seems an oddly obfuscated one. On the other hand just this part of the circuit, without the vactrol and with a coupling capacitor to remove the DC, might make an interesting distortion.