This is a simplified view to the problem:
In normal operation (Vin is positive, and Vout has reached its target value), the (high) current flows through the channels of Q2 and Q1. No current flows through the body diode of Q2, in normal operation (it does flow, during startup, as The Photon says). The reason to have two MOSFETs (instead of one MOSFET (Q1) and one Schottky diode (in place of Q2)) is exactly this one. To avoid the voltage drop that otherwise we would have across that Schottky diode.
Why does the current flow through both channels, in normal operation? Because both are on. The LT4356 uses an internal charge pump to generate a voltage Vg higher than Vgs_th for those MOSFETs (which is 2.5 V max). Imagine Vg is around 10 V (actually, between 4.5 V and 14 V). Vg1 is 10 V above Vout. So, Vgs(Q1)=10 V > 2.5 V=Vgs_th, and Q1 is on.
Q3 is on only for negative voltages below -1.4 V. So, in normal operation, Q3 is off. No current flows through the 1 Mohm resistor, and Q2 sees exactly the same Vg as Q1. So, Vg1=Vg2. How about Vgs for Q2? How much is it? Well, if Vout is (for instance) designed to be +12 V, and Vg is 10 V above it, then Vg1=Vg2=22 V. Q2 is on if its Vgs is higher than 2.5 V. For Q2 not to be on, Vin should be higher than Vg2-Vgs_th=22-2.5=19.5 V (!), which will never happen, in normal conditions. In normal conditions, Vin will be only slightly above Vout. So, Q2 is on in normal operation, and its body diode is just short circuited, contributing to zero voltage drop (which was the reason to put there a second MOSFET).
When Vin is reversed, and below -1.4 V, Q3 is on, that makes Vgs(Q2)=0, and there is no way that Q2 may conduct. Also, its body diode will be reversed biased, so it won't conduct, either. Since Q2 is in series with Q1, it does not matter what Q1 does, because no current will flow through any of them, and the load will be safe.
More: the reason for this complexity is that a silicon MOSFET is a device that can carry current in both directions, but can block only in one direction (due to the unavoidable body diode). If that body diode wasn't there, a MOSFET would be an ideal switch (able to carry and block in both directions), and a single MOSFET would be enough. Given that the diode is there, the only way to build a bidirectional-carrying bidirectional-blocking switch with them is by placing two of them in anti-series. With their gates tied together and also either a) (ideally) their sources tied together, or b) their drains tied together (as is the case, here).
GaAs MOSFETs don't have the body diode, and therefore a single device works as an ideal switch.
I assume that you have from 1 kV to 2 kV available as required for testing. If your question includes "how do I generate 1 kV" or similar please make this clar in the question.
All available data and a photo would help.
Non destructive testing may be difficult.
If you insert only the components that are relevant - eg those that cross the isolation barrier, and then apply a high voltage via suitably high resistor and monitor current, you can get a reasonably good idea of whether you have leakage.
As a system rated at 1.5 kV should protect against somewhat more than the rated limit you can do a pass/die test with 1 kV. As above you can feed the HV via a high value resistor to limit current flow if breakdown occurs. Breakdown does not have to be destructive to anything on the isolated side BUT it's very easy when you are applying 1.5 kV to have a proportion of it turn up where it shouldn't.
IF the part is from a reputable manufacturer then you are really testing your PCB. Soldering in a dummy device so that solder etc is the same in real and dummy systems and then cuttin the dummy away so only its leads are left will give you a PCB that closely resembles your finished product.
Best Answer
The whole circuit is a flyback converter so there's your first term to start googling.
The TL431 acts like a voltage comparator - as soon as the "right" output voltage appears on the isolated output, it turns the UC3842 off (via the opto-isolator) and the DC output voltage drops a tad (under load) and then (microseconds later) the TL431 signals to the UC3842 that there is an under-voltage situation. This type of feedback can be a little unstable and can lead to a bang-bang control of a sort.
The threshold at which the "regulation" occurs is when the junction of R66 and R67 is 2.5 volts (that's the reference voltage inside the TL431). If you do the math, that allows 1.042 mA through R67 and R66 - this means that the output voltage regulation point is 21.25 volts nominally.
Because a flyback converter circuit stores magnetic energy in the 1st half of the cycle without reference to the output winding, you have to use a current limiter to protect the transformer from core saturation - that is what Isense is being used for. A fixed amount of energy is stored in the primary winding and, in the 2nd half of the cycle, that energy is released from the secondary, via D2 into the output reservoir capacitor that appears to be missing in your circuit.