When is localization a quotient? When is the localization of commutative rings actually a quotient?
For instance, the germ at $x$ of the sheaf of smooth functions is the quotient of the global sections by the ideal of smooth functions vanishing on some neighborhood of $x$. But I think it is also the localization of the local sections at the maximal ideal of functions vanishing at $x$.
What are some results in commutative algebra relating localizations and quotients?
Here's an example:
Let $R$ be a local ring with maximal ideal $\frak m$ and $A$ be an $R$-module. Then $A/\mathfrak m A\cong (R/m)\otimes _R A= A_{R/\frak m}$.
 A: Let $A$ be a commutative ring and let $S$ be a multiplicatively closed subset of $A$. We want to know when $A \to A [S^{-1}]$ is surjective – for this, we can use the explicit construction of $A [S^{-1}]$ in terms of fractions.
Clearly, $A \to A [S^{-1}]$ is surjective if and only if, for every $a \in A$ and every $s \in S$, we have $\frac{a}{s} = \frac{a'}{1}$ in $A [S^{-1}]$ for some $a' \in A$. But $\frac{a}{s} = \frac{a'}{1}$ in $A [S^{-1}]$ if and only if $a s' = a' s s'$ for some $s' \in S$. Thus:

The following are equivalent:
  
  
*
  
*$A \to A [S^{-1}]$ is surjective.
  
*For every $a \in A$ and every $s \in S$, there exist $a' \in A$ and $s' \in S$ such that $a s' = a' s s'$.
  

Note that the kernel of $A \to A [S^{-1}]$ is simply $\{ a \in A : \exists s \in S . a s = 0 \}$, regardless of whether $A \to A [S^{-1}]$ is surjective.

Let's apply this in the case where $A = \mathscr{C}^\infty (M)$ for a smooth manifold $M$ and $S = \{ f \in \mathscr{C}^\infty (M) : f (p) \ne 0 \}$ for some $p \in M$.
First, observe that if $a \in A$ and $s \in S$, then there is an open subset $U \subseteq M$ such that $p \in U$ and $s$ vanishes nowhere on $U$, so $\frac{a |_U}{s |_U}$ is a well-defined smooth function on $U$. Thus, by choosing a bump function $s'$ with $s' (p) = 1$ and support contained in $U$, we can find $a' \in A$ such that $a s' = a' s s'$. Hence, $A \to A [S^{-1}]$ is surjective. 
By similar arguments, we see that the kernel $A \to A [S^{-1}]$ is the ideal of smooth functions that vanish on an open neighbourhood of $p$. On the other hand, the existence of bump functions implies that every germ of a smooth function at $p$ can be represented by a smooth function defined on all of $M$. Hence, $A [S^{-1}]$ is also isomorphic to the ring of germs of smooth functions at $p$.
A: If you don't mind assuming some finiteness conditions, there is a very nice characterization: 

Theorem: A localization $A\to A[S^{-1}]$ is a quotient by a finitely generated ideal iff there exists a decomposition $A\cong B\times C$ of $A$ as a direct product of rings such that the $B$-coordinate of every element of $S$ is a unit in $B$ and there is some element of $S$ whose $C$-coordinate is $0$, and in this case $A\to A[S^{-1}]$ can be identified with the projection $A\to B$.

In particular, if $A$ is Noetherian, this completely characterizes localizations which are quotients.  It also characterizes those quotients (by finitely generated ideals) which are localizations, since a projection $B\times C\to B$ always is a localization (take $S=\{(1,0)\}$).  This characterization is in sharp contrast with your example of germs of smooth functions, in which the ideal is not finitely generated, and the ring admits no nontrivial direct product decompositions (assuming the manifold is connected).
To prove this theorem, note first that the reverse direction is easy: if such a decomposition $A\cong B\times C$ exists, then it is easy to see that the projection $A\to B$ has the universal property of the localization $A\to A[S^{-1}]$ (any map that inverts $S$ must annihilate all elements of the form $(0,c)\in A$, since $S$ contains an element of the form $(b,0)$).
Conversely, suppose $f:A\to A[S^{-1}]$ is surjective with finitely-generated kernel $I$.  For each $i\in I$, there must be some $s\in S$ such that $si=0$.  Since $I$ is finitely generated, we can find a single $s\in S$ such that $sI=0$ (choose such an $s$ for each generator and multiply them together).  There is some $a\in A$ such that $f(a)=s^{-1}$; now consider the element $e=sa\in A$.  Since $f(e)=1$, we have $s(1-e)=0$; that is, $s=s^2a$.  Multiplying this equation by $a$, we get that $e=e^2$, so $e$ is idempotent.  Letting $B=A/(1-e)$ and $C=A/(e)$, we have a direct product decomposition $A\cong B\times C$.  
To complete the proof, it suffices to show that $I=(1-e)$; it follows that $f$ can be identified with the projection $A\to B$, and so the $B$-coordinate of every element of $S$ is a unit, and $s$ is an element of $S$ whose $C$-coordinate is $0$ (since $s(1-e)=0$).  Since $f(1-e)=0$, we have $(1-e)\subseteq I$.  Conversely, if $i\in I$, then $si=0$, so $ei=sai=0$, so $i=(1-e)i\in (1-e)$.
