$A\subset B\subset K$, $K$ field of fractions of $A$, $B$ flat over $A$ $\implies B=S^{-1}A\ ?$ Let $A\subset B\subset K$ be domains such that $K$ is the field of fractions of $A$, and $B$ is flat over $A$.
Is there a multiplicative subset $S$ of $A$ such that $B=S^{-1}A\ ?$
 A: No, flat overrings are not necessarily localizations (although the converse of course holds).  But there are important similarities, enough to motivate their being called "generalized quotient rings."
You need to generalize the notion of a localization to properly characterize flat overrings as quotient-like structures.  The appropriate generalization is to take a multiplicatively closed collection of ideals, $\mathfrak{I}$, and to define the generalized transform of $R$ with respect to $\mathfrak{I}$ as $R_{\mathfrak{I}}=\bigcup_{A\in \mathfrak{I}}A^{-1}$.  Here, by $A^{-1}$, I mean the quotient ideal $(R :_K A)$.  There's no need to limit our discussion to integral domains.  

Theorem: Given $A \subset B \subset K$, $B$ is flat over $A$ iff there exists a multiplicatively closed collection of ideals $\mathfrak{I}$ in $A$ such that $B = A_{\mathfrak{I}}$ and each $I \in \mathfrak{I}$ generates $B$ as a $B$-module. 

This is Theorem 1.3 in this paper of Arnold and Brewer.
Akiba gives an example (see beginning of section 2) showing that a flat overring need not be a localization.

Let C be a non-singular plane cubic curve defined over a field $k_0$
  and let $P$ be a generic point of C over $k_0$, and let $k$ be a field containing
  $k_0(P)$. Then the homogeneous coordinate ring $R_0=k[x, y, z]$
  of $C$ over $k$ is normal. 
Let $A=k[x, y, z]_{(x,y,z)}$, and set $\mathfrak{I} = \{\mathfrak{p}^n \mid n \in \mathbb{N} \}$ where $\mathfrak{p}$ is the homogeneous prime ideal of $A$ corresponding to $P$. Let $B= A_\mathfrak{I}$.

Then $A$ is a local domain and $B$ is a flat overring of $A$ such that no non-unit of $A$ becomes a unit in $B$.  For our purposes, this shows that $B$ is not a localization of $A$.  For Akiba's purposes, it showed the stronger statement that $B$ is not a generalized transform of $A$ with respect to any collection of invertible ideals.
