The purpose of this circuit is to create an impedance match between the signal input and the amplifier. Rs in this case is the signal resistance and the amplifier is common-gate configuration. The transistor can be sized and drain current set so that the impedance looking into the transistor source will equal Rs. That is why the problem says Rs=1/gm, you assume that you make Rs=1/gm
VnRs is the noise voltage of this signal resistance.
When calculating a noise factor, the input is grounded and the sum of the noise powers is found at the output. Or noise factor can be found by using input referred noise, it is the same answer.
I show the KCL equations on page 1. It can be tricky and of course one mistake on a sign makes the whole problem wrong, so it is best to become very good at doing this. I hope it helps.
As for the way this problem starts to find noise factor, I would never do it that way. I don't know how your teachers want you to find F, but I would draw each circuit as having a single noise source, then use superposition to add the noise sources at the output, like I have shown. Actually, I would leave out RL because it makes the problem more complicated (this does not affect F very much in practice) and find F as a sum of noise currents, but your problem includes RL so I did too.
Good luck! I know how helpful it is to see one of these problems worked out in detail. F should be unitless, so if you end up with units somewhere, you made a mistake.