The existence of an isotopy in the manifold Let $M$ be a connected closed smooth manifold and $B(1)$ the unit ball in $\mathbb R^n$. Suppose $f:M \times B(1) \to M \times \mathbb R^n$ is a smooth embedding. 
Can we find an ambient isotopy $F_t$ of $M \times \mathbb R^n$ such that $F_0=\text{id} \times \text{id}$ and $F_1 \circ f=h\times \text{id}$, where $h$ is a self-diffeomorphism of $M$. 
 A: That's not true as stated. $h$ being a diffeomorphism implies that $F_1 \circ f$ is a homotopy equivalence, thus $F_0 \circ f$ also needs to be a homotopy equivalence. One can easily construct counterexamples, such as
$$S^1 \times [-1,1] \to [-1,1] \times [-1,1] \to S^1 \times \mathbb{R}.$$
Thinking of $S^1 \subset \mathbb{C}$, the first map can be $ (z,t) \mapsto (z \cdot e^{t-1})$ and the second one $ (t,x) \mapsto (e^{it},x)$. Since the first map is contractible, the composition is contractible. Both maps are smooth embeddings so the composition is as well.
Here's a picture.

In such case you can't even hope for $h$ to be a topological embedding since that would be a homeomorphism. But in this case you can get some smooth $h$.
And of course asking $f$ to be a homotopy equivalence, or homotopic to identity (both spaces are homotopic to $M$), would yield an interesting question.
EDIT: Even requiring that $f|_{M \times \{0\}} = \mathrm{id} \times \{0\}$ (and thus $f$ is homotopic to identity) would not be sufficient. Say this time that $M=S^1$ and $n=2$, and take the map $ S^1 \times D^2 \owns (z,w) \mapsto_f (z,zw) \in S^1 \times \mathbb{C} = S^1 \times \mathbb{R}^2
$. If you view $S^1 \times \mathbb{R}^2$ as a solid torus embedded in $\mathbb{R}^3$, then $f( S^1 \times \{0\})$ and $f(S^1 \times \{1/2\})$ have linking number $1$ and no isotopy can unlink them.
So, you also need to assume at least that $Df$ as a bundle map is, in a suitable sense, homotopic through embeddings to identity. Precisely, you need to assume that there is a homotopy $f_t:M \times B(1) \to M \times \mathbb{R}^2$ between $f$ and identity and a homotopy $F_t : TM \times \mathbb{R}^n \to TM \times \mathbb{R}^n$ between $Df$ and identity of bundle maps lifting $f_t$.
At this point it sounds plausible to me, probably some $h$-principle machinery may be used to prove this.
