Is a torus a subset of $\mathbb{R}^3$ or $\mathbb{R}^4$?

These are two different embeddings of the same topological space, i.e. the surface of the bagel in $\mathbb R^3$ and the product of two circles in $\mathbb R^4$ are homeomorphic. Note that they are not isometric though; the torus embeded in $\mathbb R^4$ as a product of circles is flat, while the surface of the bagel has regions of positive curvature and of negative curvature.

Neither! A torus is an abstract topological space defined, as you say, to be the product $S^1 \times S^1$. Since, as it happens, $S^1 \subset \mathbb{R}^2$, we get a natural embedding of "the" torus into $\mathbb{R}^4$. However, there are of course other ways of representing this space; one can change coordinates, apply functions to it, and so on. And among these alterations are some that "project" the torus down into $\mathbb{R}^3$, where it resembles a bagel.

This is a little bit of an aberration, since not every two-dimensional manifold can be embedded in $\mathbb{R}^3$, the Klein bottle being a counterexample (also, this is related to your question in that it is more or less a twisted torus). And more generally, an $n$-dimensional manifold can be embedded in $\mathbb{R}^{2n}$ but not necessarily lower. So if you insist on thinking of a manifold as "being" a subset of some $\mathbb{R}^N$, that's the one to choose.