Well I have a doubt about a rank $k$ vector bundle. My definition of vector bundle is: A rank $k$ vector bundle is a triple $(\pi, E, M)$ where $E$ and $M$ are smooth manifolds and $\pi:E\rightarrow M$ is a smooth submersion, which satisfies:
(i) for every $p\in M$ the fiber $F_p=\pi^{-1}(p)$ is a $k$-vector space.
(ii) given $p\in M$ there is an open set $U\subseteq M$ containing $p$ and a diffeomorphism $\phi:\pi^{-1}(U)\rightarrow U\times \mathbb R^k$ such that $\textrm{pr}_1\circ \phi=\pi$.
(iii) For every $q\in U$ the map $\phi_q:E_q\rightarrow \mathbb R^k$, $e\mapsto (\textrm{pr}_2\circ \phi)(e)$, is a linear isomorphism..
My doubt is: If I have $\pi^{-1}(U)=E$ and $U=M$ in $(ii)$ and I consider $\phi^{-1}:M\times \mathbb R^k\rightarrow E$ can I say $\phi^{-1}(p, \cdot)=\phi_p^{-1}$?
You didn't state it in your definition, but $\phi_p\colon E_p\to\{p\}\times U$ is the restriction of $\phi$ to the fiber $E_p=\pi^{-1}(p)\subseteq\pi^{-1}(U).$ Hence the identity you asked about is certainly satisfied, since in general the inverse of a restriction of any function is the restriction of the inverse.