We know with feedback, the voltages at pins 9 and 10 of UC1c have to be the same. The output will give whatever voltage it needs to make sure pin 9 adjusts to the voltage at pin10 (0.2V). The output ramps up until the bottom right diode in the bridge conducts. This passes a voltage to the bottom of R8. Current flows UP through R8 and through the top left diode of the bridge. This current then flows on through R9 and V2.
If you think current instead of voltage: the same current flows through R8 as R9 and V2. If R8 is 10x the value of (R9 + V2), you will get 10x the voltage across it.
Put another way, with feedback, we know the voltage at pin 9, so hence know the current flowing through (R9 +V2). If R8 is 10x bigger, with the same current, the voltage will be 10x bigger.
C3 smoothes the waveform into dc.
the next stage is a differential to single ended converter that maps the capacitors voltage to a ground referenced voltage to feed into the ADC.
Damn good interview question I think....
If you want the rest of the circuit explaining, let me know and I will talk you through it
(I have a sinewave osc kit to cater for the front end if it helps)
Best of luck