Charge Injection Confusion
I'd say it's a lot simpler if you just stick with the standard electrical engineering convention of voltage, current, and charge instead of trying to think in terms of electrons (and/or "holes"). That's how the standard electrical engineering curriculum teaches it. So we just build up from simple fundamentals:


Electric current is just the flow of charge from point A to point B.
A capacitor builds up voltage as it accumulates charge, as defined by Q=CV.
A MOSFET has a small amount of parasitic capacitance between all terminals( Reference)


Showing the above circuit with the parasitic capacitance added:







simulate this circuit &ndash; Schematic created using CircuitLab


We start with Vin at a constant 1V and CK high (let's say 5V). The MOSFET M1 is ON, so Cds is shorted out and both Cgd and Cgs have +4V across them. CH has +1V across it, and Vout is +1V as well.
When CK goes low (0V) then M1 turns off. We can ignore Cgd, since there is no way for it to source or sink current to CH. We can also ignore Cds, since it is both a very small capacitance and a very small voltage across the capacitor (0.00 - 0.04V). But Cgs is the important one. Pulling CK down to 0V means we must introduce a voltage of -1V across Cgs, versus the +4V it had before.


That's the key point. The capacitor starts out with Q1 = Cgs*4V and then must consume enough charge to re-stabilize at Q2 = Cgs*(-1V), so it draws a total charge of (Q2-Q1) out of CH. Since CH is much larger than Cgs, this 5V change on Cgs corresponds to a 0.04V change across CH.

In the circuit shown by Razavi, there is another parasitic capacitance introduced which has a complementary voltage across it, so any excess charge introduced by M1 would be consumed by M2, and vice versa, leaving the net charge on CH unchanged.