Unique extendable functions... Is there a theory? Motivation: Take a continuous function $f$ from a topological space $X$ to a hausdorff space $Y$. If $g$ is a continuous function from $X$ to $Y$ that coincides with $f$ in a dense set, then $g=f$.
Take an analytic function $f: \mathbb{C} \rightarrow \mathbb{C}$. If $g$ is an analytic function that coincides with $f$ in a set with limit point, then $g=f$.
Okay, now given the motivation, my question is the following: 
Is there a theory which has the objective of characterizing functions that satisfy a certain "unique extension" property, in a general context?
For example, we could define:

Definition: Given a family $\mathcal{F} $ of maps $f:X \rightarrow Y$ , $f \in \mathcal{F}$ is said to satisfy $P$-unique-extendability on $(X,Y)$ if for every subset $A \subset X$ which satisfies the property $P$, we have that every function $g \in \mathcal{F}$ which coincides with $f$ in $A$ is such that $f=g$.

Therefore, for example, every continuous function is dense-unique-extendable on $(\mathbb{R}$, $\mathbb{R})$
Maybe there is a categorical POV of this?
 A: This is strongly connected to the notion of an epimorphism. In the category of Hausdorff spaces, the epimorphisms are those continuous maps with dense image. In the category of smooth manifolds, the epimorphisms have the same description; but this can be simplified because of the identity theorem.
A: If we assume your class of morphisms is closed under composition, contains identities, and inclusions of subobjects you're interested in, so that we can place all this into a single category, then properties of extensions translate to elementary properties of the $\mathrm{Hom}$ functor.
Let $f : A → B$ be a morphism (an inclusion of a subspace $A ⊆ X$ you mentioned would be a motivating example) and say that morphisms from $B$ to $C$ are uniquely determined on $f$ if all morphisms $g$, $h : A → C$ that agree on $f$ (ie. $gf = hf$) are in fact equal. In other words, this is exactly saying that $\mathrm{Hom}(f, C)$ is an injective function, so we're basically asking when is this the case.
As Martin Brandenburg notes, it is obviously true if $f$ is an epimorphism (more abstractly, hom functors preserve monos, so the contravariant one translates epis to injections). Since you are primarily interested in subobjects, in your case $f$ would in fact be a bimorphism. This of course is not necessary: even if morphisms to $C$ are determined on $f$, morphisms to $D$ needn't be (in other words: hom-functors don't necessarily reflect monos). One way to ensure that epicness is necessary is to demand that $C$ be a cogenerator, ie. that for every pair $f$, $g : X → Y$ of distinct parallel morphisms there exists a morphism $h : Y → C$ that distinguishes them ($hg ≠ hf$). An example of a cogenerator is the unit interval for functionally Hausdorff or better spaces. This is awfully restrictive, but I don't think there's much more you can say in such a specific situation where both $f$ and $C$ are fixed.
Of course if you allow $C$ to vary then you get exactly the definition of an epimorphism.
ps. I changed your "uniquely extendable through f" to "uniquely determined on f", because I think "unique extendability" should imply extendability, which is not the case here:
every ring morphism $\mathbb Q → \mathbb Z/(5)$ is trivially uniquely determined on $\mathbb Z → \mathbb Q$, but no morphism $\mathbb Z → \mathbb Z/(5)$ can be extended to $\mathbb Q$.
Note that extendability also has an easy categorical formulation: saying that every morphism $g : A → C$ extends through $f : A → B$ is exactly asking for the function $\mathrm{Hom}(f, C)$ to be surjective. Unique extendability would then correspond to that function being bijective. Existence of extensions is a more difficult problem because in general hom functors neither preserve nor reflect epis. Two ways to ensure the extensions exist would be demanding that $f$ be a split mono, so that $A$ is a retract of $B$, or that $C$ be an injective object, but both of these are overkills (although the first one is exactly what you need if you allow $C$ to be arbitrary).
