Of course a local ring is semilocal.
If R is local, and B is a ring with an image of R embedded in B, and B is a finitely generated R module, then B is semilocal.
Let M be the maximal ideal of R, and let W be a maximal ideal of B. Let K be the quotient field B/W.
Let M act on B, as an R module. This is the span of elements in M times elements in B, and is an R submodule of B, and an ideal in B.
Suppose M*B is not contained in W. Mod out by W, and B/W, or K, becomes an R module. We can let R, or M, act on K. By assumption, M*B does not lie in W, hence M*K is a nonzero submodule of K.
A homomorphism takes R into B, possibly an embedding, and another homomorphism takes B onto K. Combine these, and R acts on K from the left, and respects right multiplication by K. Thus the submodule R*K, or M*K, is an ideal in K. The only nonzero ideal in K is K, thus M*K = K.
Now K, the quotient of B, is a finitely generated R module. Since R is local M is the jacobson radical. Thus jac(R) maps K onto K, and by nakiama's lemma, K = 0. This is a contradiction, hence M*B lies in W for every maximal ideal W.
If we are interested in the maximal ideals of B, we can look at the maximal ideals of B mod M*B. Call this ring V. It is both a ring and an R module. Since M carries B into M*B, V is also an R/M module.
Let F be the field R/M. Now V is an F module, or an F vector space. In fact V is a finite dimensional F vector space, since B is a finitely generated R module.
The kernel of the map from F into V is an ideal, namely 0, and F embeds in V. Every ideal in V can be scaled by F, hence every ideal is an F vector space, with a well defined dimension. Since V is finite over F, infinite chains are not possible, and V is noetherian and artinian.
An artinian ring has finitely many maximal ideals, hence V is semilocal, and B is semilocal.
The product of the maximal ideals is contained in their intersection, and is 0. Also, maximal ideals are pairwise coprime. Apply the chinese remainder theorem, and B is isomorphic to the direct product of rings B/Wi. Each of these component rings is a field. Therefore a semillocal ring, mod its jacobson radical, is a finite direct product of fields.
Now return to the case where R is local, and maps into B, and B is a finitely generated R module. We already showed B is semilocal.
Mod out by the jacobson radical and the quotient is a ring, and an R module, and the direct product of fields.
Next, mod out by the maximal ideal W. This is a ring homomorphism, and an R homomorphism, and it selects one of the fields in the direct product, rather like a projection. Let K be the resulting field.
Remember that M carries B into W, hence M kills K. Thus K is an R/M module. Let F = R/M, and F maps into K. The kernel of this map is an ideal in F, and it has to be 0, hence F embeds in K.
Since B is finitely generated over R, K is finitely generated over F. In other words, K is a finite field extension of F, and this holds for each K across the direct product. Note, these fields need not be isomorphic to F or to each other, as shown below.
The prime ideal generated by 13 has two prime ideals above it, generated by 3+2i and 3-2i. Think of Z13 as a local ring. (It happens to be a field.) The gaussian integers mod 13 becomes a Z13 module, spanned by 1 and i. It is finitely generated. It must be semilocal, and the residue fields must be field extensions of Z13. In fact there are two such fields, mod 3+2i and mod 3-2i, and each is isomorphic to Z13.
There is but one prime lying over 19, namely 19. The resulting semiloca ring is already local, and the residue field has order 361.
A somewhat simpler example maps Z7 into the field of order 49 cross the field of order 343. The two residue fields need not be isomorphic to each other, or to F; yet each is a finite extension of F.