Electronic – the ROF pin on the power supply

What current will be used is determined by the load, not the power supply. We've had a number of questions like this before, like "how do I limit the current of my 5A supply to 1A".

What your device will need is just the addition of the current for the different components on the board. You usually start with the power-hungry parts: motors, relays. For a microcontroller you'll have to look up in the datasheet what it needs at a given clock frequency.

If you have a lab supply which displays current you may use that as a rule-of-thumb, but you may want to give it some headroom. Used current is definitely useful if you want to design a power supply.

L1, Bead, C5, C2: You are guessing right, these components act as a filter. The transients caused by U1's switching need to be kept away from the mains input. While you can read the diagram from left to right when thinking about the direction of the power flowing from the input via the rectifier to the transformer, the switching noise has its source at the node labeled "C" (U1.1, U1.2). One way for the noise to leave the device is from right to left, via the rectifier towards the input, and the filter is right in this path.

R2, R5: These are called start-up resistors. A small current through R2 and R5 charges C3 and C6, and likely, an internal Z-Diode from U1.4 to U1.5 holds the voltage across C3 and C6 at 10...15 V. Once C3 and C6 are charged, U1 is able to start and from there, D6 provides the auxiliary power to C3 and C6 and for U1. R2 and R5 do not act as a filter (as long as you don't get too theoretical and call R2+R5 and C3+C6 a low pass RC filter, which would of course be true but not very practical). Actually, there even is a resistor to dampen the filter, but this is the one they've put parallel to L1 (R10).

If you want to understand the circuit, it'll help a great deal to redraw it with the major internal components and building blocks of U1, and not sticking to the IC's pinout when drawing the schematic.