Yes that's a particularly confusing piece of text. Your analysis is basically correct:
1) With S1 closed and S2 open, the capacitor charges or discharges until the voltage across it is Vin, and the charge is C1 * Vin. (they've called this qin)
2) With S1 open and S2 open, and an ideal capacitor the capacitor's charge does not disipate and the voltage remains the same.
3) With S1 open and S2 closed, the capacitor charges or discharges until the voltage acorss it is Vout, and the charge is C1 * Vout (they've called this qout).
If you do the sequence of 1 then 2 then 3, then 1 again.
During step 3, the charge on the capacitor starts at qin and ends up as qout
so during that step a charge of qin-qout is transferred to the output
During step 1, the charge on the capacitor starts at qout and ends up as qin
so during that step a charge of qout-qin is transferred from the capacitor to the input, or more sanely a charge of qin-qout is transferred from the input to the capacitor.
So after one complete cycle a charge of (qin-qout) is transferred from the input to the output.
plugging in the values you get qin-qout = C1*(Vin-Vout).
As an aside the problem with this sort of design is that you'll end up dissipating a lot of energy in resistance of the wires and switches, due to equipartition as the charge flows in and out to/from the decoupling capacitors on Vin and Vout. Most uses of this sort of capacitor switching power supply are used where Vin is equal to Vout, either to turn a +ve voltage into a -ve voltage (the switches turn the capacitor upside down), or as a doubler (the switches take the capacitor and stack it ontop of the supply voltage). Both types can be seen in the popular MAX232 series of line drivers.