Here are some sufficient conditions:
Let $A$ be a local domain, $K$ its fraction field, $L$ a finite extension of $K$, and $B$ the integral closure of $A$ in $L$. If $A$ is Noetherian and integrally closed, and $L$ is separable over $K$, then $B$ is necessarily finite over $A$, and so is semi-local (by going-up/down-type theorems).
If $A$ is not integrally closed, but its integral closure in $K$ is finite over itself, then again $B$ will be finite over $A$.
The condition that the integral closure of $A$ in $K$ be finite over $A$ is (at least by some people) called N-1. More generally, the condition that $B$ be finite over $A$ is called N-2, or Japanese. (The letter N here stands for Nagata, and I believe Grothendieck coined the adjecive Japanese for these rings because these properties were studied by Nagata and the commutative algebra school around him in Japan.)
So if $A$ is a Japanese ring, then $B$ will be finite over $A$, and hence semilocal. Of course, this is rather tautological: its utility follows from the fact that many rings (indeed, in some sense, most rings --- i.e. most of the rings that come up in algebraic number theory and algebraic geometry) are Japanese. E.g. all finitely generated algebras over a field, or over $\mathbb Z$, or over a complete local ring, are Japanese.
Here are some useful wikipedia entries related to this topic: Nagata rings and Excellent rings.
Let $a_1,\ldots,a_n$ generate $A$ as a $\mathbf{Z}$-module, and let $F=\mathbf{Q}(a_1,\ldots,a_n)\subseteq K$. Note that $\mathbf{Q}(a_1,\ldots,a_n)$ contains $A$, so in fact we must have $\mathbf{Q}(a_1,\ldots,a_n)=K$. Now, because $A$ is finite over $\mathbf{Z}$, it is in particular integral over $\mathbf{Z}$, which means that every element of $A$ is the root of a monic polynomial with coefficients in $\mathbf{Z}$. In particular, this is true of the $a_i$, and so each $a_i$ is algebraic over $\mathbf{Q}$ (integrality over a field is the same as algebraicity). So $K$ is generated over $\mathbf{Q}$ by finitely many algebraic elements, and therefore is a finite extension of $\mathbf{Q}$.
Best Answer
The following proof is based on the assumption that $A$ is Noetherian and is integrally closed (in its field of fractions $K$):
(1) If $L$ is a finite separable extension of $K$, then the bilinear form $(x,y)\to \text{Tr}_{L/K}(xy)$ on $L$ is nondegenerate. (Exercise)
(2) Choose a basis of $L$ over $K$. We can assume, after possibly scaling by elements of $K$, that this basis consists entirely of elements of $B$ (the integral closure of $A$ in $L$). (Exercise) Let $(v_1,\dots,v_n)$ be this basis of $L$ (as a vector space over $K$).
(3) We know by (1) that there is a basis $(w_1,\dots,w_n)$ of $L$ over $K$ "dual" to the basis $(v_1,\dots,v_n)$ of $L$ over $K$. More precisely, there is a basis $(w_1,\dots,w_n)$ of $L$ over $K$ such that $\text{Tr}_{L/K}(v_iw_j)=\delta_{ij}$ for all $1\leq i,j\leq n$ (where $\delta_{ij}$ denotes the Kronecker delta). (Exercise)
(4) Let $x\in B$ and write $x=\Sigma_{j=1}^n x_jw_j$ where $x_j\in K$ for all $1\leq j\leq n$. Note that $xv_i\in B$ for each $1\leq i\leq n$ and thus $\text{Tr}_{L/K}(xv_i)\in A$. (Exercise) However, $\text{Tr}_{L/K}(xv_i)=\Sigma_{j=1}^n x_j\text{Tr}_{L/K}(w_jv_i)= x_i$. Therefore, $x_i\in A$ for all $1\leq i\leq n$ and $x\in \Sigma_{i=1}^n Av_i$.
(5) Since $A$ is Noetherian, it follows that $B$ is a finitely generated $A$-module. (Exercise)
An interesting corollary that is fundamental in algebraic number theory:
Corollary Let $A$ be a Dedekind domain and let $K$ be its field of fractions. If $L$ is a finite separable extension of $K$, then the integral closure of $A$ in $L$ is also a Dedekind domain.
Proof. We know that $B$ is Noetherian since it is a finitely generated $A$-module (and $A$ is Noetherian). Also, $B$ is integrally closed in $L$. (Exercise) Therefore, all that remains to show is that every non-zero prime ideal of $B$ is maximal. (Exercise) Q.E.D.
I hope this helps!