The power factor is managed ("corrected" is really the wrong term, although its the common one) by making the current follow the voltage. In your schematic, the bus voltage will be a bit higher than the peaks for the AC waveform. The inductor, FETs, diode, and capacitor form a boost converter. This converter takes the rectified AC input voltage and makes the bus voltage.
If the control system only regulated the output voltage, there would be no PFC happening. What it does instead is regulate the average current thru the diode to be proportional to the instantaneous rectified AC input voltage. Remember that the ideal load from a power factor point of view has the current in phase with the voltage. Another way of looking at it is that load on the AC line needs to look resistive. Just like a real resistor, you want to keep the current proportional to the voltage.
Of course that is at odds with regulating the bus voltage. This is handled by having a fast response to the AC input voltage but a much slower response to regulating the bus voltage. In other words, the AC line still sees a resistance, but the resistance value is slowly changed as needed to keep the bus voltage near its target value.
You can check out my Digital PFC Control writeup for more background on PFC and a way I came up with to keep the current proportional to the voltage without having to measure the current. I've got a patent on that, which also includes using digital computation to control the bus voltage more accurately. With a little computational power, you can know what ripple is caused on the bus due to following the AC line voltage, then use that to determine what changed due to varying demand from the load. This allows adjusting to load changes more quickly than the conventional approach but without defeating the PFC function.
Power factor correction of a linear inductive load is exactly the same as tuning a parallel circuit of a capacitor and inductor. You pick a capacitor that works with the value of the transformer's magnetizing inductance to satisfy this: -
60Hz = \$\dfrac{1}{2\Pi\sqrt{L_M.C}}\$
Where \$L_M\$ is the transformer's magnetizing inductance and C is the capacitor chosen to "neutralize" the current taken by the coil.
For a pure lossless inductance and capacitance, the resultant current taken from the supply is zero.
If your transformer primary indicates an inductive reactance at 60Hz of 300 ohms, the magnetizing inductance value is this divided by \$2\Pi\times 60\$ = 0.8 henries.
The capacitive reactance required is 300 ohms and this is 8.8 uF. As a sanity check: -
F = \$\dfrac{1}{2\Pi\sqrt{0.8\times 8.8\times 10^{-6}}}\$ = 59.98Hz
Best Answer
Assuming the load is inductive, then shunt capacitors will improve the load power factor. However, with a stand-alone generator it's rarely necessary, and there are several considerations.
Do Not Do This on the output of a square wave or 'modified sine' inverter. Shunt capacitors will overload the inverter on the fast edges.
Power factor correction doesn't save you any power (fuel consumption) in the generator unless the transmission lines to the load have a significant loss compared to the load. It's not worth adding shunt C for the fun of it.
If you do decide to do it, then under-correct rather than over-correct, don't try to hit nominal spot-on.
Mount the shunt capacitors with their loads so that the shunt C changes when the load changes. It can be dangerous to add shunt caps to a generator and then run it off load, due to voltage rise with its residual inductance.