Do smooth embeddings preserve quotient bundles? Let $M$ be a smooth connected oriented $d$-dimensional manifold. Let $S \subseteq M$ be a $k$-dimensional embedded submanifold which is also compact, connected and orientable.
Suppose we are given a smooth embedding $F:S \to \mathbb{R}^d$.

Question:
Suppose that $TM|_S$ is a trivial vector bundle. Is it true that the quotients $TM|_S \big/ TS$ and $S \times \mathbb{R}^d\big/dF(TS)$ are isomorphic as vector bundles?

The triviality of $TM|_S$ is an obvious necessary condition, since if $TM|_S \big/ TS \cong S \times \mathbb{R}^d\big/dF(TS)$, then $$TM|_S \cong TS \oplus TM|_S \big/ TS \cong dF(TS) \oplus S \times \mathbb{R}^d\big/dF(TS) \cong S \times \mathbb{R}^d.$$
Of course, isomorphic bundles and subbundles can induce non-isomorphic quotients, so $TS \cong dF(TS)$ and $TM|_S \cong S \times \mathbb{R}^d$ do not imply the quotients are isomorphic.
My guess is that the answer can be negative in general, but I don't know how to find a counterexample.
Note that when the codimension $d-k=1$, the answer is positive:
The quotients are isomorphic.
 A: The answer to your question is no. As a counter example, consider $S=S^2$, the $2$-dimensional sphere, and $M=TS$, the total space of the tangent bundle. The inclusion $S\hookrightarrow M$ we think of is the usual one, that is, we identify $S$ with the zero section.
Why is this a counter example?
Fact $1$: The normal bundle of $S$ in $M$ is non-trivial. Indeed, this normal bundle is just the tangent bundle itself, which is non-trivial.
Fact $2$: The sphere $S$ can be embedded in $\mathbb{R}^4$ with a trivial normal bundle. Indeed, the standard embedding of $S$ in $\mathbb{R}^3$ as the unit sphere has a trivial normal bundle, and so, the composition $$S\hookrightarrow\mathbb{R}^3\hookrightarrow\mathbb{R}^4,$$where the right hand embedding can be any linear one, has a trivial normal bundle too.
The only thing we need now is 
Fact $3$: The restricted bundle $TM|_S$ is trivial. To see this, note first that $$TM|_S=TS\oplus TS.$$Now, every rank $4$ bundle over the $2$-sphere is determined by a loop in $SO(4)$ (every such bundle is trivial on both the northern and southern hemispheres, and can thus be described by a transition map along the equator). But the fundamental group of $SO(4)$ is isomorphic to $\mathbb{Z}/2$, and so, every loop in $SO(4)$ of the form $\alpha*\alpha$ is null-homotopic, where $\alpha$ is any other loop in $SO(4)$. 
Let $\beta\in\pi_1(SO(2))$ denote the homotopy class corresponding to the tangent bundle $TS$. Then under an isomorphism $\pi_1(SO(2))\to\mathbb{Z}$, the class $\beta$ is mapped to $\pm2$ (this is a known fact), which means that there is an $\alpha\in\pi_1(SO(2))$ such that $\beta=\alpha*\alpha$. 
In conclusion, the homotopy class in $\pi_1(SO(4))$ which corresponds to $TS\oplus TS$ is $$\beta\oplus\beta=(\alpha\oplus\alpha)*(\alpha\oplus\alpha)=0,$$verifying Fact $3$.
Edit: In fact, I believe the direct sum $E\oplus E$ is trivial for any rank $2$ bundle over the $2$-sphere. This can be shown using arguments similar to the ones above.
Another edit: I wonder what the answer would be if you required $M$ to be Euclidean too. In other words, we know that in general, a manifold can be embedded in Euclidean space in non-homotopic ways (think of knot theory, for example). But can a manifold be embedded in Euclidean space in two different ways that yield different normal bundles?
