This problem is somewhere in the domain of combinatorics - and it's hard to specify much beyond that (and "combinatorics" is very broad), because it doesn't fit neatly into some smaller field of study. The most relevant mathematical objects are really sets and functions (for really elementary things), directed graphs (if we just use brute force), and orders (if we look deeper). In fact, it's not so hard to completely draw out the theory here.
To describe the question mathematically: You have some set $P$ of positions on the objects and a set of objects $O$, each of which is some subset of $P$ (the subset it covers). You also have some set of colors $C$. Every "state" of the game is represented as some function $f:P\rightarrow C$ giving the colors of each position, and legal transitions are choosing some subset $X$ of objects and some color $c$ and changing the value of $f$ on every point $p$ not covered by something in $X$ to the color $c$.
You could represent the set of states and transitions as a directed graph, but this doesn't give great insight - better is to study the definition a bit and reduce it to something more manageable.
If you think about this definition more rigorously, but without reference to any external theorems, you can describe a strategy that always works. In particular, there's a relation hiding here:
A point $p$ is at least as specific as a point $q$ if every object that covers $q$ also covers $p$.
This is equivalent to saying that $p$ is at least as specific as $q$ if it is not possible to cover $q$ without also covering $p$. Equivalently, if $q$ is not covered by some set of objects, $p$ is not covered either. This is really important because if we wished to color them different colors, then we'd have to color $q$ first, then color $p$, since otherwise we could not color $q$ without overwriting the color of $p$.
Let's denote this relation by writing $p \leq q$, referring to how the smallest region including $q$ that can be left uncovered must, under this definition, contain the smallest region including $p$ that can be left uncovered. This is transitive meaning that if $p \leq q$ and $q\leq r$ then $p\leq r$ and it is symmetric meaning that $p\leq p$. This defines a structure known as a preorder. There is an associated idea of congruence given by $p\sim q$ if $p\leq q$ and $q\leq p$ - which, in elementary terms, means that $p$ and $q$ are covered by exactly the same set of objects.
We can then prove a theorem, with an explicit description of the strategy (assuming finitely many objects):
A coloring $f:P\rightarrow C$ can be achieved if and only if $p\sim q$ implies $f(p)=f(q)$.
Which basically says that only the obvious constraint applies. To prove this, we can first look at the equivalence classes of $P$ under $\sim$ - which basically means that we can either pick one point $p$ to represent each region defined as "the set of points covered by this set of objects and no others" or we can consider each such region as its own object. This reduces us to having only finitely many elements to work with; for notation, the set of such regions is denoted $P/\sim$ and each of them, due to the hypothesis, has some desired color valid over that whole region.
However, once we have chosen these representatives, the relation $\leq$ becomes a partial order - meaning that if $p\leq q$ and $q\leq p$ then $p=q$. Then the strategy becomes easiest:
Let $S$ be the set of regions in $P/\sim$ that are not the correct color and choose a maximal $x\in S$ (i.e. some $x$ so that if $y\in S$ and $x\leq y$ then $x=y$). Cover the ball with every object not covering $x$ and color that region with the color desired for $x$. Repeat this until you finish.
The proof of correctness is easy: the set of points in $P/\sim$ which are at least as specific as one which is incorrectly colored gets smaller each time we apply this move.
Then, we can translate back into simple terms:
Look at all the points that are not yet the correct color. Of these points, find one which could be covered by the largest number of objects. Cover the ball with all the objects not covering that point and color that point with the color it ultimately wishes to be. Repeat until you win.
This is guaranteed to work on any puzzle that is not blatantly impossible (i.e. has two points, covered by the same set of objects, but of differing desired colors).
Note that this solution only really grazes any branch of mathematics: we borrow the terminology of order theory to solve and visualize things, but we didn't use any substantive theorem other than "every finite partially ordered set (poset) has a maximal element" - and we wouldn't have known that orders would be relevant just looking at the question. Basically, this is less of a "go study this specific field" problem and more of a "have a broad view of mathematics and get your hands dirty" problem.