Notice that / is heavily overloaded. Sometimes it means a ring or field extension, sometimes it means a quotient ring, and sometimes it means a fraction ring. Sorry for the confusion. I guess you just have to know what / means by context.
Now where were we? The rings S and R contain the ideals H and G respectively, and the quotient ring S/H is an integral extension of the quotient ring R/G. Verify the following steps.
Since both rings contain T/T, or 1, S/T is a ring extension of R/T. We will show S/T is an integral extension of R/T.
If the numerator of a fraction is x, write p(x) = 0, where p is monic, with coefficients from R. A ring homomorphism carries S into S/T. Here x is mapped to x/1, or xw/w if T does not contain 1. In any case, we can apply the homomorphism to x, and to the coefficients of p(x). Thus x/1 ∈ S/T is integral over R/T. Multiply by 1/j ∈ R/T, and any x/j is integral over R/T. Thus S/T is an integral extension of R/T.
Assume R is a field. Given x in S-R, write p(x) = 0, where p is monic. Suppose p is the product of two smaller polynomials. When evaluated at x, the result is 0, and since S is an integral domain, x is a root of one of the two smaller polynomials. Therefore we can assume p is irreducible. This represents a field extension of R, hence x is invertible.
Conversely assume S is a field. Let x be a nonzero element of R and let y be its inverse in S. Write p(y) = 0 where p is monic. Multiply through by xn-1, and y lies in R after all.
Suppose H is a prime ideal. This means xy in H implies x is in H or y is in H. This certainly holds when we restrict to R, so G is prime in R. The quotient rings S/H and R/G are integral domains, and by the above, one is a field iff the other is a field. thus, if H is maximal in S, H∩R is maximal in R.
Now assume G is prime in R. We want to lift this to a prime ideal H lying over G.
Let T be the set R-G, which is multiplicatively closed. We showed above that S/T is an integral extension of R/T.
Select a maximal ideal in S/T and let H be the numerators of this ideal. Thus H is a prime ideal in S, and H/T gives the same maximal ideal back again. See ideal correspondence for more details.
Let C = H∩R. If a fraction in R/T and another fraction in H/T represent the same class, establish a common denominator d, and the first numerator is still in R, and the second is still in H. To equate these fractions, write d(a-b) = 0. Since 0 is in H, a prime ideal, and since d is nonzero, a = b. The numerator is in both R and H, hence in C. Conversely, every fraction in C/T is in H/T and R/T. Intersection and localization commute.
Recall that C/T has to be prime, and since H/T is maximal, C/T is also maximal. However, G/T is the largest ideal in R/T. This because RG is a local ring. So G/T is the maximal ideal, and G/T = C/T. Thuse C = G, and the prime ideal H intersects R in G. We have found a prime ideal H lying over G.
Using our earlier result, H is maximal iff G is maximal. There are primes over primes, and maximals over maximals.
Intersection with R implements a map from prime ideals in S onto prime ideals in R. For a fixed element x in R, a prime G in R misses x iff its preimage H in S misses x. Base open sets pull back to base open sets, and contraction to R implements a continuous function from spec S onto spec R.