Let $(\Phi, \phi):(E, \pi_E, M)\longrightarrow (F, \pi_F, N)$ be a morphism between two smooth vector bundles. I know there is an isomorphism of $C^\infty(M)$-modules $$u: C^\infty(M)\otimes_{C^\infty(N)} \Gamma(F)\longrightarrow \Gamma(\phi^*(F)), f\otimes \alpha\longmapsto f\cdot (\alpha\circ \phi, \textrm{id}_M),$$ where $(\alpha\circ \phi, \textrm{id}_M)$ is the smooth section of $\phi^*(F)$ given by $$(\alpha\circ \phi, \textrm{id}_M): M\longrightarrow \phi^*(F), p\longmapsto (\alpha(\phi(p)), p),$$ and where $\cdot$ stands for the $C^\infty(M)$-module multiplication on $\Gamma(\phi^*(F))$.
What is the inverse of this isomorphism?
Thanks
Remark. It is easily seen that $$\Gamma(\phi^*(F))=\{(\alpha, \textrm{id}_M): \alpha\in C^\infty(M, F); \pi_F\circ \alpha=\phi\}.$$ So, given $(\alpha, \textrm{id}_M)\in \Gamma(\phi^*(F))$ we must produce an element of $C^\infty(M)\otimes_{C^\infty(M)} \Gamma(F)$. I guess we should take $1\otimes \overline{\alpha}$ where $1: M\longrightarrow \mathbb R$, $x\longmapsto 1_{\mathbb R}$, and $$\overline{\alpha}: N\longrightarrow F, q\longmapsto \alpha(p), p\in \phi^{-1}(q).$$ Notice $\overline{\alpha}$ (if well defined) lives in $\Gamma(F)$ for $$\pi_F\circ \overline{\alpha}(q)=\pi_F\circ \alpha(p)=\phi(p)=q.$$ The problem boils down to showing $\overline{\alpha}$ is well defined.