The other side of the 'input' capacitor is pulled down to a much lower voltage, which means the charge prefers to flow into it. (And then this alternates so the next time the one it is transferring charge to is pulled lower)
From what I can find, it's something like 99.999% of charge moves, but it varies from CCD to CCD and can cause artifacts in bigger ones if it's too low. I couldn't find a great indication of what affects it, other than defects in the crystal lattice causing charge trapping, which apparently tends to be an issue in aging telescopes, especially space ones like Hubble.
OK but what I mean is that if you tie two capacitors together, and one of them has a terminal at a much lower potential, then unless that voltage is minus infinity, then the other one will not be completely discharged.
Capacitors store a certain amount of charge, in that regard like a small battery. The voltage on the positive side is relative to the negative side.
If you have a 5V supply with respect to ground (your reference voltage), and you connect a capacitor between it and ground, it will charge to 5V.
If you then disconnect the capacitor and connect the negative side to the 5V supply, then the positive side will still be 5V relative to the negative side of the capacitor, but the positive side will be 10V with respect to ground.
Now take a second capacitor. Let's call the positive sides of the two capacitors as p1 and p2 respectively, and the negative sides for n1 and n2 respectively.
If you then connect a second capacitor in parallel with the first, so both n1 and n2 are connected to 5V, the charge will equalize between them. If you then connect n1 to ground, then p1 will be negative relative to p2 and charge will start flowing into capacitor 1 from capacitor 2.
Using transistors as digital switches, you can do this "dance" of alternating where you connect the negative sides to and thus move the charge along.
