You are right, the circuit is drawn wrong.
https://www.eleccircuit.com/wp-content/uploads/2008/09/compaq-power-supply-200w.jpg
is better.
Looking at the circuit, it seems more likely that the OP will be able to run the charger on 220 V
In that, when the 110/220 V switch is closed, C6 is charged through one diode in the bridge when the supply is +ve and C5 is charged through the other diode when the supply is -ve. Two of the diodes are not used. The voltage is around √2 * 110 on each capacitor, so around 155 V on each and 310 V total.
When the 110/220 V switch is open, the series combination of C5 & C6 is charged by both halves of the cycle, using all four diodes, two in one direction and two in the other. R1 and R2 keep C5 and C6 at very similar voltages. The voltage is around √2 * 220, which is also 310 V.
On the original circuit, I can't see any balancing resistors, and I can't see what the capacitor arrangement is. However, having two capacitors, a link and four diodes means that it is quite possible that the circuit is similar to the computer power supply one, and cutting the link is all that it needed. The circuit must be checked first or it could all end in smoke.