Looking at the right side of the opto, there are two different circuits
One is a 10V power supply. This consists of D6 (as a rectifier), R3, R4, and R5 as a current limiter, D1 as a regulator, and C1 and C2 as storage caps. It is used to provide power so the optoisolator can turn on the FET.
Current in R3, R4, and R5 is ((1.414 * 230)-10)/(33k+33k+220R) = 4.8mA
Dissipation in R3 or R4 is 4.8ma*4.8mA*33k = 634mW so you would use a 1W resistor for these. As for why they used R3 and R4 in series, it's to limit the voltage across either. This way, you will get about 158V across each instead of 316V across one. R5 is to limit inrush when C2 first starts to charge. Note that C1 must be big enough to keep the power on even during times when the FET is on (effectively shorting out the input to the power supply section)
The other circuit is the switch. It consists of the transistor side of the optoisolator, the FET, and bridge. For this part of the circuit, don't look at the bridge as a device to convert AC to DC. Look at it as diodes.
When the FET is off, no current will flow between the two AC terminals as each pair of diodes D2, D3 and D4, D5 contains two diodes in opposite direction.
When the FET is on, think of it as a short between the positive and negative terminals of the bridge (that's close enough; it really is a low value resistor). When this happens, current will flow from the right AC input, through D2, the FET, and D5 back to the load during one half cycle, and D3, the FET, and D4 the other half cycle. In both cycles, however, the current flowing through the FET itself will always be in the same direction.
The optoisolator keeps the high voltage on the right side of the circuit separate from the low voltage stuff on the left. That includes your computer (if hooked up to the PIC with a programmer or serial port), and you (if there are buttons or controls hooked up to it). It, and the proper PC board layout, is there for safety.