An ascending or descending chain of prime ideals in S contracts to an ascending or descending chain of prime ideals in R. If not, then a prime P in R has two primes Q1 and Q2 in S, lying over P, with one containing the other.
What about chains that go beyond infinity? If the union of an ascending chain of prime ideals in S is prime, or is contained in a larger prime ideal, this prime ideal in S contracts to a prime ideal in R that is the union of, or contains the union of, the foregoing prime ideals in R. A similar result holds for the intersection of a descending chain of prime ideals. Use transfinite induction to show that ascending or descending chains of prime ideals (beyond infinity) in S, contract to same in R.
This does not imply chains in R lift to chains in S. Yes, P1 might contain P2 in R, and there exists Q1 and Q2 lying over P1 and P2, but that doesn't mean Q1 contains Q2.
Localize about P, so that RP becomes a local ring. The primes in S, lying over P, persist in the ring SP, and remain distinct. Furthermore, they lie over a maximal ideal in the local ring, namely PP, hence they are maximal. A finitely generated module over a local ring is semilocal, and has finitely many maximal ideals. Hence the fiber of P is finite.