> what if you put the switch before the caps?
Works exactly the same.
> wouldn't they build up less of a charge if they're not connected to the circuit when not being used?
It is the circuit which charges them up. One side sits near 3V DC, the other side near 1V DC. The cap holds a 2V charge.
Doesn't matter which "side the switch is".
If it was just the one cap, there is a 2V "pop" at turn-on and turn-off, but these are masked by the whole circuit coming up from zero to 9V and back.
Here we have one cap charged to 2V, and five more un-charged. When switched-in, these caps must charge-up to 2V. This kicks right into the input, and is large compared to guitar signal, POP!
Meanwhile a cap which has been charged to 2V and then switched out self-discharges to zero. On a good cap this should be "slow". I bet if you switch through all the caps, and then switch back, the pops are much less the second time. However if you play a while, they discharge back to zero, and will give a 2V pop when put back in.
The 1Meg-5Meg resistors across the switch keep the "idle ends" near the same DC voltage they will have in use. Pop is much less. Assuming 100Meg capacitor leakage and 1Meg bleed resistors, the pop is about 1/100 of before, a big reduction.
There's other ways. You can redesign the circuit so both ends of the caps are the same DC voltage. But for a simple system like this, that's a major effort and surely more costly.
> have the caps wired in series
In this case, that reduces the pop by roughly the number of caps. All-in, the 2V divides across all caps, at the other limit only one cap has the 2V. You can add bleeders on all but the last cap, then it takes the whole 2V at all times and the others are at zero VDC whether bled or shorted. Saves one resistor. However I -think- the series-string is liable to need more total nFd than the parallel, so the cost may be the same. Even if I'm wrong, the cap-math for series is too tedious for one-off DIY construction.