Is it true that $\mathfrak{X}(M\times I)\simeq (C^\infty(M\times I)\otimes_{C^\infty(M)} \mathfrak{X}(M))\oplus C^\infty(M\times I)?$ Let $M$ be a smooth manifold and $I:=
[0, 1]$. Let $\mathfrak{X}(M)$ be the $C^\infty(M)$-module of vector fields on $M$ and $\mathfrak{X}(M\times I)$ be the $C^\infty(M\times I)$-module of vector fields on $M\times I$. Is it true that there is an isomorphism of $C^\infty(M\times I)$-module $$\mathfrak{X}(M\times I)\simeq (C^\infty(M\times I)\otimes_{C^\infty(M)} \mathfrak{X}(M))\oplus C^\infty(M\times I)?$$
The $C^\infty(M)$-module structure on $C^\infty(M\times I)$ is induced by the morphism of algebras $\textrm{pr}_1^*:C^\infty(M)\longrightarrow C^\infty(M\times I)$ where $\textrm{pr}_1:M\times I\longrightarrow M$ is the projection on the first component and the $C^\infty(M\times I)$-module structure on $C^\infty(M\times I)$ is given by the pointwise product of functions.
Thanks.
 A: This is how I would do it. It is not really a complete answer to the question, but it is far too long for a comment.
Suppose the tangent bundle of $M$ is trivial, i.e. it is isomorphic to $M\times\mathbb{R}^n$ where $n$ is the dimension of $M$. Then it admits a global basis, which we denote by $\partial_1,\ldots,\partial_n$. The tangent bundle of $I$ is canonically trivial and has basis $\partial_t$, and $T(M\times I)\cong TM\oplus TI$ via the projection maps. Therefore, a vector field $X\in\mathfrak{X}(M\times I)$ can always be written as
$$X(m,t) = X^1(m,t)\partial_1+\cdots+X^n(m,t)\partial_n + X^t(m,t)\partial_t$$
for functions $X^i\in C^\infty(M\times I)$. It follows that we have an isomorphism
$$\mathfrak{X}(M\times I)\cong C^\infty(M\times I)\otimes_{\mathbb{R}}\mathbb{R}^{n+1}\ .$$
In particular, this is always true if $M$ is an open subset of Euclidean space, and thus it is always true locally for any manifold.
More generally, we can do the following. A vector field $X$ on $M\times I$ can be decomposed into $X=X^M + X^I$ with $X^M:M\times I\to TM$ and $X^M:M\times I\to TI$. Since the tangent bundle of $I$ is trivial, we can see $X^I$ as a smooth map,
$$X^I:M\times I\longrightarrow\mathbb{R}$$
so that we have an isomorphism
$$\mathfrak{X}(M\times I)\cong\Gamma(M\times I,TM)\oplus C^\infty(M\times I)$$
where $\Gamma(M\times I,TM)$ denotes the maps $M\times I\to TM$ such that the image of $(m,t)$ is in $T_mM$. i don't know if more can be said in general.
