Quotient rule, differential operator on a localization is well-defined, underlying geometry? 
Using the quotient rule, we obtain that the notion of differential operator on a localization is well-defined:$$\mathcal{D}_A(B_f) \cong \mathcal{D}_A(B)_f.$$Here, $B$ is a commutative $A$-algebra, $\mathcal{D}_A(B)$ denotes the ring of differential operators of $B$ over $A$, and $B_f$ is $B$ localized about the multiplicative set generated by $f$, where $f \in B$.

Is there a good way to intuit/picture this geometrically?
 A: Elaborating Allen Knutson's answer a little:
On smooth manifolds, a linear operator (between spaces of functions, or of sections of a vector bundle) are local (support non-increasing) if and only if it is a differential operator (Peetre's theorem).
Does Peetre's theorem hold also in the algebraic setting? Is an operator which prolongs to each localization, necessarily a differential operator?
A: Ring of differential operators are generated by their first filtration components (differential operators of order one) that can be split into functions and vector fields.
Localization is just an algebraic way of restriction to open subsets, i.e. if $B$ is the ring of functions on a variety $X$, then $B_f$ is the ring of functions on $X \setminus Z(f)$, where $Z(f)$ is the zero set of $f$. 
If I have a vector field on $X$ I can always restrict it to open subset $X\setminus Z(f)$ (i.e. $ \mathcal{D}_A(B)_f  \subseteq \mathcal{D}_A(B_f) $) On the other hand, if I have a vector field on $X\setminus Z(f)$ I can want to view it as a restriction of a vector field on $X$. The possibility to do this, I think, depends on which kind of functions you consider.
