Hi,
I feel u r missing one thing here.Let me explain the complete scenario 2 u.
First only the feed back wire was connected between i/p and o/p.So Vo=Vx=V+=0
(virtual gnd concept )b'cos of -ve feed back(with no attunuation).Hence charge across Cin=0 in sampling mode.
In the amplification mode the feed back is hindered with a capacitor.Hence now there might b some diff. betn 2 i/p nodes.And Vx is no longer =0(line 6).
So where from this new charge across Cin come from?Definitely it has 2 come from the node Vin.So charge across Cin = Cin*Vx.But the basic poperty of cap is it doesn't allow the voltage 2 change across it.Hence the charge stored in Cin will also appear on the +ve terminal of Ch.Hence the new charge across Ch is now Ch*Vo+Cin*Vx.
I think this helps.
regards,