1. You say "if the right side of the cap is open", but it isn't really, is it? After all, there's that other diode and a (possibly uncharged) capacitor downstream of it: doesn't this constitute a load?
Yes, of course. I left the cap open to try to simplify the discussion about how the cap works. The load will obviously affect this charge and discharge. With a load the capacitor will discharge until it has transferred all the charge it can.
For example if it were connected to a second equal value cap C2 to ground that started at 0V, then the result with 10V applied at the left side would be that the transfer of charge would leave the first cap with 0V across it and C2 would have +10V across it.
If we repeat the process with out discharging C2 then you end up with +15V across C2. Repeating the process many times incrementally increases the voltage on C2 until it approaches 20V (thus a voltage doubler).
2. Here's my problem still: so we have the first capacitor charged by the first negative half-cycle thus:
-(0V) ---||--- + 10V
So if the positive half-cycle "lifts" both sides of the capacitor equally, wouldn't we then have
+10V ---||--- + 20V
(because we added 10V to both sides)? giving a net charge of 10V, not 20V.
Obviously, it doesn't work this way. But it looks as if it works this way, from the explanations I've read.
Yes that is correct. The net voltage on the cap is still 10V, but the output voltage is 20V.
Looking forward to having the errors in my thinking corrected.