I'm starting to read about smooth vector bundles. My notes suggest that instead of giving a local trivialisation at each point, it is sufficient to give a local frame.
I can see that given a rank $r$ vector bundle $(E, \pi)$ over $M$ and a local frame $\{s_i: 1 \leq i \leq r\}$ for $E$ over $U \subseteq M$, we can define a local trivialisation $t$ by setting for $v_q \in E_q = \pi^{-1}(q)$, $$t(v_q) = (q, \alpha_1, \dots, \alpha_r) \label{eq}\tag{1}$$ where $v_q = \sum_{i=1}^r \alpha_i s_i(q)$. The only non-trivial part of seeing that $t$ is really a trivialisation is seeing that it is smooth. An argument for this assuming that $(E, \pi)$ already has the structure of a vector bundle is given here.
My understanding of what is written in my notes suggests that this is also true without assuming that $(E, \pi)$ is already a vector bundle. That is, we should have something like
Proposition: Let $\pi:E \to M$ be a smooth map of manifolds such that for each $p \in M$, $E_p = \pi^{-1}(p)$ is an $r$-dimensional vector space. Suppose further that for each $p \in M$, there is an open neighbourhood $U$ of $p$ and a smooth frame $s_1, \dots, s_r: U \to E$. Then $(1)$ defines a trivialisation of $E$ over $U$. In particular, $E$ is a rank $r$ vector bundle over $M$.
It is clear that we can use the argument given in the linked question as long as for each $p \in M$, $E_p$ is contained in an open subset of $E$ that is diffeomorphic to $U \times \mathbb{R}^r$ where $U$ is an open neighbourhood of $p$ in $M$ (by looking at a suitable restriction of a local frame at $p$). However it is not clear to me that such a subset of $E$ should exist without assuming that some trivialisation already exists.
I'd like to know if my proposition is true and if so how this attempt at a proof may be fixed (or if there is some entirely different argument that works).
As it stands, the Proposition is wrong because the vector space structure may not vary smoothly.
Let's start with $\pi:E = U \times \mathbb{R}^n \rightarrow U$ and the trivial sections ($U$ a manifold). Now pick some $p\in U$ and change the vector space structure on $E_p$ such that $0_p$ changes the position and $s_1(p),...,s_n(p)$ remains a basis. Give the newly obtained object (i.e. manifold + projection + new vector space structure at $p$) the name $E'$, then according to the proposition $E'$ should be a vector bundle as well. But then $0':U\rightarrow E'$ (the new zero section) is not smooth, which is impossible for vector bundles.
Example: Consider the projection $\pi:\mathbb{R}^2\rightarrow \mathbb{R}$ onto the first factor as a smooth map between manifolds. For $p\neq 0$ equip the fibre $\pi^{-1}(p)=\{p\}\times \mathbb{R}\subset \mathbb{R}^2$ with the vector space structure inherited from $\mathbb{R}^2$.
In the fibre $V=\pi^{-1}(0)=\{0\}\times \mathbb{R}$ we choose a weird vector space structure such that $0_V = (0,1).$ To this end define addition in $V$ by $$ (0,a) +_V (0,b) = a+b-1$$ and scalar multiplication by $$ \lambda \cdot_V (0,a)= (0,1+\lambda(a-1)).$$
Next consider $s:\mathbb{R}\rightarrow \mathbb{R}^2,p\mapsto(p,2)$. This is a section of $\pi$ (i.e. $\pi\circ s = \mathrm{id}$) and also constitutes a frame, because for each $p\in \mathbb{R}$, the vector $s(p)=(p,2) \in \pi^{-1}(p)$ is different from $ 0_{\pi^{-1}(p)}$ and thus a basis.
Now we are in the situation of your Proposition: $\{s\}$ is a smooth frame (even a global one), nevertheless $\pi$ is not a vector bundle. The reason for this is a again that the zero section $$0_\pi:\mathbb{R}\rightarrow \mathbb{R}^2, p \mapsto \begin{cases} (p,0) & p\neq 0 \\ (0,1) & p= 0 \end{cases}$$ is discontinuous.