127v/220v AC sensor - What's the role of this particular resistor?
The first resistor is working with the capacitor as an RC filter. Since you are sensing AC at 50~60Hz, you need this RC filter to get a constant output instead of a pulsed output. Remember your otpo-coupler has a LED (diode) inside, so it will only produce light when voltage is positive.

The second resistor is there to limit the current from your AC 220Vrms (which is 311V peak) so that is within the opto-coupler's LED forward current range. The voltage between your LED´s pins will always be it´s forward voltage (1.2V to 1.4V), so the resistor will bring down the 311V to 1.2 to 1.4V. Without this resistor, you would have 311V between the LED pins, and this would fry the poor thing.

The equivalent input circuit is something like this:







simulate this circuit &ndash; Schematic created using CircuitLab

Ok so how did they figure out the 220kOhms value?
First, take the worst case scenario: 220Vrms AC gives us 311Vpeak, and LED forward voltage is 1.2V.
So the voltage between the 311V and the diode is 311 - 1.2V = 309,8V  

OHM Law: R = V/I        (R=resistance, V=voltage, I=current)

In this case, the current chosen for the LED was about 1.4mA or 0.0014A, so R = 309.8V/0.0014A

R = 221k Ohms. The closest we get to this is 220kOhms

With 127Vrms (179.6Vpeak), your LED current will be: 
I = 178.4V/220000 = 811uA

I have simplified the calculations, because this is a simple case, but keep in mind that real life engineering calculations will be more complex than this.