Applying these transformations in either order need not reproduce the original ideal. The contraction of the extension always contains the original ideal, while the extension of the contraction is contained in the original ideal.
If the image of a commutative ring R lies in the center of S, the contraction of a prime/semiprime ideal is same. Let P be a prime ideal in S, and pull it back to R. Since 1 maps to 1, the preimage of P is a proper ideal. If xy lies in the preimage, then f(x)*f(y) lies in P. Remember that f(x) and f(y) commute with everything in S, so The product of the principal ideals generated by f(x) and f(y) lies in P, either f(x) or f(y) lies in P, and either x or y lies in the preimage of P. This is our test for primality in a commutative ring. Similar reasoning pulls a semiprime ideal back to a semiprime preimage.
This does not hold for noncommutative rings. Let R be the integer polynomials over x and y, with x2y2 mapped to 0. Since x2y2 = 0, 0 is not a prime ideal. Build S by adjoining w to R, such that w and R do not commute. Given any two polynomials p and q in S, p*w*q is nonzero. Just look at the lowest degree term of the product. By default, 0 is a prime ideal in S. It contracts to 0 in R, which is not prime. This embedding also violates semiprime contractions.
Map an ideal H into another ring S, and the extension is the linear combinations of all the elements of H mapped into S. Do this for H1 and H2 and consider the product ideal in S. It is spanned by pairwise products from the two extensions. Each pairwise product is a linear combination of images of elements from H1 times a linear combination of images of elements of H2. This reduces to a linear combination of images of pairwise products from H1 cross H2, which is the extension of H1*H2 in S.