Q:

A quotient field is finitely generated as a module?

When is the following true and why?

If a commutative ring $A$ is a finitely generated module over a commutative ring $B$, then its quotient field $K(A)$ is a finitely generated module over $K(B)$.

A:

It is always true that $K(A)$ is finitely generated as a module over $K(B)$ .  
Since you mention quotient fields, we may  assume that $A$  is a domain.
We may also assume that $B\subset A$ and thus that $B$ is a domain too: if this is not the case and you have an algebra $\phi:B'\to A$, just consider $A$ as a $B=\phi(B')$-algebra.   
With these preliminaries out of the way, let $S=B\setminus \lbrace 0\rbrace$.
Then (by base change or by hand) $S^{-1}A$ is a finitely generated $S^{-1}B$-module .
  Since $S^{-1}B$ is a field and $S^{-1}A$ is a domain,  $S^{-1}A$ is a field too by the Useful Lemma below.
Finally,we have  $A\subset S^{-1}A\subset K(A)$ and since $S^{-1}A$ is already a field, we have $S^{-1}A=K(A)$ : we have proved that  $K(A)=S^{-1}A$ is a finitely generated module (=finite-dimensional vector space ) over the field $K(B)=S^{-1}B$. 
Useful Lemma:
If  the commutative domain $R$ is a finite-dimensional algebra  over a field $K$, then $R$ is a field.
Proof:
Let $0\neq r\in R$. The multiplication map $m_r:R\to R:x\mapsto rx$ is an injective endomorphism because $R$ is a domain, hence it is surjective by elementary linear algebra .
Thus there exists $s\in R$ with $m_r(s)=rs=1$ and thus $r$ is invertible: $r^{-1}=s$.
NB I have simplified my original proof by invoking  the Useful Lemma instead of  integral extensions.

A:

I would just like to add to Georges' answer. He mentions that if $A$ is finitely generated as a $B$ - module, then $S^{-1}A$ will be finitely generated as an $S^{-1}B$ module where $S = A - \{0\}$. I would like to mention that this result can be proved as follows:
Recall we have the $S^{-1}B$ - module isomorphism
$$S^{-1}A \cong  S^{-1}B \otimes_B A$$ 
 such that every element in the tensor product can be written as an elementary tensor of the form $\frac{1}{s} \otimes a$ for some $s\in S$ and $a \in B$. But then $A$ was finitely generated as a $B$ - module so we can write $a = \sum_{i=1}^n a_ib_i$ for some $b_i \in B$ and $a_i's $ the generators of $A$. Hence we can write $a$ as 
\begin{eqnarray*} \frac{1}{s} \otimes a &=& \frac{1}{s} \otimes \sum_{i=1}^n a_ib_i \\
&=& \sum_{i=1}^n \frac{b_i}{s} \otimes a_i \\
&=& \sum_{i=1}^n  \frac{b_i}{s} \left(\frac{1}{1} \otimes a_i\right) \end{eqnarray*}
showing that $S^{-1}A$ is finitely generated as an $S^{-1}B$ - module. 
$$\hspace{6in} \square$$
Appended for the OP: Suppose we have integral domains $A,B$ with $A \subseteq B$ and $B$ integral over $A$. Then $A$ being a field implies that $B$ is a field.
Proof: Take any $y \in B$ and by assumption $y$ satisfies an integral dependence relation
$$y^n + a_{n-1}y^n + \ldots a_0 = 0$$
with the $a_i \in A$. Then you can rewrite this as 
$$(y^{n-1} + a_{n-1}y^{n-2} + \ldots + a_1)y = -a_0.$$
Now $a_0 \neq 0$ for otherwise this contradicts $B$ being an integral domain. It follows that we can divide through by $-a_0$ to get that 
$$(\text{some stuff})y = 1$$
where the stuff is the inverse of $y$, consequently $B$ is a field.

