The output of a PI controller is a slow moving voltage, in response to the error signal input. It will settle slowly and respond slowly to set-point changes, and it will react slowly to transient changes. It won't be bang-bang against the rails. The 2 diodes back-to-back before it gets to the feedback input is just a diode OR, to make sure that only one of the 2 controllers is in control at any given time (Voltage or Current).

The internal reference is still being used, and the voltage from the PI controllers (whichever one is in control) will do whatever it needs to do, and reach whatever voltage it needs to reach, within it's abilities, in order to make the reference input voltage = 1.235 Volts. So, working backwards from the reference input, you see that Vref is 1.235 V. Then there is the voltage divider, so the top of the voltage divider will be Vref * 2 = 1.235*2 = 2.470. Then there is a diode drop to the output of either PI controller, which means the PI controller output wants to stay at 2.470+0.2V = 2.67 V. With error feedback from the OUT+ and OUT- points into the first AMP U2A (which is a weighted difference amp), and the PWM setpoint from the micro going into the PI controller's setpoint input, this will be what determines the output path as it settles to 2.67 V.

The output voltage is measured at OUT+ relative to OUT- , not at OUT+ relative to GND. So the controller can reach 0 Volts output because it's not actually at 0 Volts relative to GND, it's going to be at 0 Volts relative to OUT-. So as long as OUT- and OUT+ are EQUAL, it's considered 0 Volts output.

I didn't do the math, but you can analyze the U2A difference amplifier using supposition to get the transfer function, and determine what value is required for OUT+ equal to OUT- that gives Vout = 0 for that amplifier. Note that R13, R14 and R11 inject a 50mV voltage source into the amplifier too. So your supposition calculations will have 3 voltage sources, 50mV, OUT+ and OUT-. Supposition means set all voltage sources to 0V except 1, calculate the transfer function for 1 voltage source only, then repeat for the others. The final transfer function is the sum of the 3 transfer functions. When you figure out the final transfer function you can set OUT+ equal to OUT- and simplify it further. Then solve for Vout = 0 Volts. This 0 volts is what will be matched with the setpoint PWM signal, and I am assuming that 0 volts PWM = 0 volts output desired. Once solved, you will probably find that be OUT+ = OUT- = 1.235 V :-) ( or maybe slightly higher )..