Will love to understand how the circuit works based on this schematics
It is powered from +/-24V and there are regulators for +/- 12V on top of the schematic.

U4 is wired as non-inverting amp with a gain of 2. 

We got 8 CPC1918 OptoMOS relays, presumably activated by sending enough voltage into the "PADx" inputs to light the internal LED. These select feedback resistors between 10R and 10Meg (R6,8,10...). So we can just look at one of the feedback paths, for example the one with 10R resistor, and pretend the others don't exist.

U4 is inside U2's feedback loop, it is a high current opamp, and there is a 10R 50W resistor, this means that U4 is used as a high current buffer with a gain of 2. If we take it out of the picture for a moment, the resulting schematic is a transimpedance amplifier:



In this circuit, the opamp adjusts its output to hold its "-" input at the same potential as its "+" input (0V), and if we consider the input bias current to be zero, then I1 flows into Rf, thus the output voltage is I1 * Rf. The opamp's negative input acts like a virtual ground, able to sink and source current, as long as the opamp is able to adjust its output voltage to make sure I1 flows through Rf.

Your schematic is the same thing, with an extra "reference electrode" input which allows the user to choose at which voltage the other input should be held, and it has selectable Rf via the OptoMOS relays.

It is a selectable gain transimpedance amplifier, able to process quite high currents (a few amps as per LM1875). The voltage output is in the lower right corner of the schematic.