The same question comes to me today. And my answer and understanding are:
In the circuit, the left capacitor is C1, connected to positive pole; the right one is C2, connected to negative.
After fully charged, we have +Q1 and -Q1 on two sides of Capacitor C1, and +Q2 & -Q2 on C2. Same quantity of opposite charges ensure there's no net electric flux around C1 & C2, from Gauss Theorem. Actually, at this equivalence stage, there's no electric field within the wires, only electric field within the capacitors.
Now suppose Q1 not equal to Q2. Using Gauss theorem respectively for C1 and C2, there's still no net electric flux. Then look at the part with one plate of C1 & C2 connecting via the wire, i.e. |-------| . Since -Q1+Q2 not vanishes, the net electric flux not vanishes, leading us to different electric fields E1 and E2 within capacitors.
(-Q1+Q2)/epsilon=-E1*A+E2*A =(-U1/d1 +U2/d2)*A= (-U1*C1/epsilon +U2*C2/epsilon)=(-Q1+Q2)/epsilon
where E1&E2 are electric fields, U1&U2 are voltages, d1&d2 are capacitors' separations, A is the plate area which is supposed to be equal for two capacitors.
Therefore that Q1 not equal to Q2 does not go against any physics rules. It's only the result of reaching the fully charged equivalence, together with E1=E2.
So why should the same amount of charges when equivalence? I understand as the usage of a battery in a circuit. Battery affects the free electrons to move, so there're extra electrons with -Q, also holes with +Q. Here C2 is connected to the negative pole of battery, so -Q2 is firstly accumulated on right plate of C2, along with Q1 on left plate of C1. There'are electric fields everywhere currently, and we got Q2, -Q1 accumulated under such electric fields. Yet in this process, |+Q2| not equal to |-Q2|; on the contrast, |+Q1| always equal to |-Q2|. The free electrons move under electric fields, until |+Q2| equaling to |-Q2|, the so-called equivalence condition. And electric fields within the wire also disappeared, they finished their job of "transporting electrons".