Your two questions are "Why are P(xR)= xR and P(xN)= 0?" and "what is the meaning of (I- P)(x)?".
You are projecting x onto R. That is, xR= P(x), the component of x lying along R. If some vector, y, is already pointing in the R direction, P(y)= y. If you project again, since xR is already along R, projecting again just gives you the same thing again.
As for P(xN)= 0, that is only true if N is orthogonal to R. Your picture does not make it look like that but you say "ker P= N". If xN is in ker P, then by definition of kernel, P(x)= 0. Are you sure you aren't told somewhere that R and N are orthogonal?
Finally, (I- P)x= Ix- P(x)= x-P(x). It is x, with P(x) subtracted. If you drop a perpendicular from the tip of x to R, the intersection is at the tip of R- you have a right triangle with hypotenuse x, "near side" P(x), and opposite side x- P(x). If you "follow" the path from the origin of x to its tip, then down that perpendicular (call it "y"), you get to the tip of P(x): x+ y= P(x) so y= P(x)- x= -(I- P)x: -(I- P)x and so (I-P)x is always perpendicular to R.
