Does any homeomorphism between these two manifolds factor through a higher-dimensional space? As the answers to this question (I think) successfully argue, it is often more useful to consider manifolds as intrinsically, i.e. without reference to an ambient Euclidean space. I am wondering if I have found a simple example of this phenomenon.
Consider the torus trefoil knot:  
It can be embedded into $\mathbb{R}^3$, as the above pictures hopefully illustrate. What is perhaps surprising, however, is that it is in fact homeomorphic to the regular torus in $\mathbb{R}^3$ (I think).

Question: Is it impossible to "unknot" the trefoil torus into a regular torus without leaving $\mathbb{R}^3$? I.e. does any homeomorphism between the two "factor through a higher dimensional space"?
Does this impossibility (if it is true) show the need for abstract manifolds? I.e. is this a concrete example where ambient spaces actually make things more confusing, rather than less?
The only homeomorphisms I can think of which "take place in $\mathbb{R}^3$" involve self-intersections, which, being "unphysical", I think implies factoring through a higher dimensional space. The same might be true for cutting the knot, "untying it", and then reattaching it at the same two places it was cut.
 A: Your question doesn't quite make sense. The torus, $T_1$, as a manifold, and the trefoil, $T_2$, as a manifold, are homeomorphic because there's a bicontinuous map between them. 
On the other hand, the pair $(R^3, T_1)$ and $(R^3, T_2)$ are not homeomorphic as pairs: there's no bicontinuous map from $R^3$ to $R^3$ whose restriction to $T_1$ gives a bicontinuous map from $T_1$ to $T_2$. (This requires some proof, of course.) 
You can also think of $Q = S^1 \times S^1$ as a manifold; the torus $T_1$ that you've drawn above is the image of $Q$ under an embedding (a locally maximal-rank injective map $f_1$ from $Q$ to $R^3$), as is $T_2$ (with map $f_2$. Asking whether you can "untie" the torus amounts to asking if there's a homotopy, i.e., a continuous map
$$
H: Q \times [0, 1] \to \mathbb R^3
$$
with the property that $H(p, 0) = f_1(p)$ and $H(p, 1) = f_2(p)$, and the property that $p \mapsto H(p, t)$ is an embedding for every $t$ and that $H$ is $C^1$ smooth (which prevents the so-called "bachelor's isotopy). There is no such homotopy (which again requires proof). [Alternative; you can ask for an ambient isotopy...and again no such thing exists.]
If you regard $\mathbb R^3$ as being in $\mathbb R^4$, though, you can ask whether there's a homotopy from $Q \times [0, 1]$ to $\mathbb R^4$ which looks like $f_1$ and $f_2$ at its two ends and is smooth and for each fixed $t$ is an embedding; the answer to that is "yes". So in that sense, you can untie a trefoil in 4-space, but not in 3-space. 
