First, one important point: people who study quiver varieties seem not to usually take stack quotients, but rather GIT quotients (though there ARE very good reasons to do this if you like geometric representation theory) which leads them to come up different dimension formulae from you. The reason is that you are think of a fine moduli space, whose stack dimension is the geometric dimension of the underlying variety minus the dimension of the automorphism group of the object, and modules over an algebra ALWAYS have automorphisms (if nothing else, they have multiplication by constants).
Now, the answer proper: You seem to have mixed up two of the more popular notions of a quiver variety (leading your confusing dialogue with Greg in the comments), and Kevin seems to have mixed in a third, which may or may not actually be the one you had in mind (all of which are, of course, closely related).
If you have two vertices and one arrow, then there are two things you can do.
You can take the moduli space of quiver representations of the path algebra of that quiver, which is given by $\mathrm{Hom}(V,W)/GL(V)\times GL(W)$. This has a very small dimension $(nm-n^2-m^2)$ and should probably be thought of as finitely many points (indeed, this quiver only has finitely many representations of any given dimension. The indecomposibles look like $k\to 0$, $0\to k$ and $k\to k$), all of which have a bunch of automorphisms.
Now it sounds like what you intended to do was take the "hyperkählerization" of this quiver variety. What you should do for this is take the cotangent bundle of $\mathrm{Hom}(V,W)$, but before you mod out, you have to impose a moment map condition. The reason this is a good idea is that you want a resulting variety which is a holomorphic symplectic result (just like the cotangent bundle), which you can also think of as hyperkähler by picking a hermitian metric on $\mathrm{Hom}(V,W)$. This moment map condition is basically that both possible compositions of maps along your arrows are 0 (note: I think this is not a flat map, so I think if you want to really think properly about this story, you should probably take the derived fiber). Then take the quotient of that.
What Kevin is referring to is probably the most popular definition of quiver varieties for geometric representation theory. This is yet another definition, where he interpreted one of your vertices as a shadow vertex. This extra layer of confusion results from some (IMHO poor) notational choices of Nakajima.
Let $\mathcal O$ be the structure sheaf of $\mathbb P^1$. Then $\mathcal O \oplus \mathcal O(1)$ is rigid and generates the derived category of coherent sheaves on $\mathbb P^1$. Thus, it is a tilting object, and so the derived category is equivalent to the category of modules over its endomorphism ring, which is the path algebra of the Kronecker quiver.
For $\mathbb P^n$, you need $n+1$ objects to generate the derived category. You can take $\mathcal O(i)$ for $0\leq i \leq n$; this will be a tilting object again, but its endomorphism ring will not be hereditary; you will get a quiver with relations. This shouldn't be surprising, though: path algebras of quivers have global dimension one, so you shouldn't expect their derived categories to agree with derived categories of sheaves on higher-dimensional varieties.
Best Answer
This is more or less a longer version of Phil's comment. Unfortunately I don't know definitive references, so there might be some technical errors below. A possible reference is Keller's paper.
Here's the most high level statement: for any (pretriangulated, which just means formally add all cones, shifts, and sums) dg-category $C$ (there is a natural way to enrich $\mathsf{Coh}(X)$ as a dg-category) with a chosen set of compact generators $E_\alpha$ (in the sense of a weak triangulated generator of the category of compact objects), letting $Q$ be the associated full dg-category with these generators as objects, we have that the dg derived category of $Q$-modules is dg-equivalent to $C$. The functor $C \rightarrow Q\text{-mod}$ is $\operatorname{Hom}(E_\alpha,-)$. Note all $\operatorname{Hom}$ I write are derived.
Fixing this choice of generators $E_\alpha$, we get something "like" a quiver (a category "is like" a quiver, and a module over a category "is like" a representation) but is not a quiver in the classical sense. In particular (warning -- the further down on this list you go, the further out of my depth I go):