Mumford's proof that O_X(X_f)=R_f intuition?

I'm having some trouble understanding "why" his proof works: I've re-read this proof in Mumford's Red Book several times and tried sketching out a picture, and I sort of get it, but I still find it kind of "magical" and don't know how one would be motivated to use this approach. It goes something like this: O_X is the structure sheaf of irreducible variety X and X_f is the distinguished open {x\in X: f(x)\neq 0}. It's straightforward to check that O_X(X_f)\supset R_f. To prove the subclaim that O_X(X_f)\subset R_f , we let F be a member of O_X(X_f), which is defined in this context as a subring of the field of fractions of R as \bigcap_{x\in X_f} O_x (O_x = {f/g: f, g\in R, g(x)\neq 0} being the stalk at x), where R is the coordinate ring of X. Then (where I feel like a rabbit was pulled out of a hat), define the ideal B={g\in R: gF\in R}. We want to prove F is a member of R_f by showing that f^n\in B for some positive integer n. If x\in X_f, there's some juggling of quantifiers and eventually one concludes that F=h/g where g(x)\neq 0, from which one finds that g is an element of B such that g(x)\neq 0. From this, we see that the vanishing set of B, V(B), must be a subset of V(f)={x: f(x)=0}, and applying the Nullstellensatz, we get f\in rad(B), as desired. I guess I don't really see geometrically what's going on here; it just seems like a trick, followed by carefully reasoning about the which points/regular functions contain/are contained in what. Could someone here explain what's going on in this proof? Also, how does one extend this to reducible varieties? (It's still true, I think?) My understanding is that you can't define the structure sheaf in this way (as an intersection of O_x in the field of fractions) because of zero divisors in the coordinate ring. Sorry if these are too elementary/boring questions! I asked r/learnmath without getting a helpful reply (other than to ask r/math).