Does a covering by trivializations imply existence of a linearly compatible one, for a smooth $\mathbb{R}^k$ fiber bundle?

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Let $\pi : E\rightarrow B$ be a surjective submersion between smooth submanifolds. Let $k=\mathrm{dim}\,\mathrm{ker}\,\pi_*$.

We define a local trivialization of $\pi$ with fiber $\mathbb{R}^k$ to be a pair $(U,\phi)$ where $U\subset B$ is open and $\phi : \pi^{-1}(U) \xrightarrow{\approx} U\times \mathbb{R}^k$ is a diffeomorphism such that $\phi(y) \in \{\pi(y)\} \times \mathbb{R}^k \;\forall y\in \pi^{-1}(U)$.

We define an atlas of trivializations to be a collection $\{(U_\alpha,\phi_\alpha)\}_\alpha$ where each $(U_\alpha,\phi_\alpha)$ is a local trivialization and the $U_\alpha$'s cover $B$.

We say an atlas of trivializations is linearly compatible if for every indices $\alpha,\beta$, for every $p \in U_\alpha \cap U_\beta$, the map $\mathbb{R}^k \rightarrow \{p\} \times \mathbb{R}^k : v \mapsto \phi_\alpha(\phi_\beta^{-1}(p,v))$ followed by $\{p\} \times \mathbb{R}^k \rightarrow \mathbb{R}^k : (p,w) \mapsto w$ is $\mathbb{R}$-linear.

My question is, if $\pi$ admits an atlas of trivializations with fiber $\mathbb{R}^k$, then does it also admit a linearly compatible atlas?


Edit: as has been pointed out in the comments, this fails if we relax $E$ and $B$ to topological manifolds and "smooth"/"diffeo" to "continuous"/"homeo". I've now changed the question to focus on the smooth case.