@TedShifrin I don't deny that this gives you a vector bundle. But I don't think that would be a good definition yeah.
But that's also different, the transition functiosn express how you go from one fiber to another, if you regard (real) vector spaces as vector bundles over a one point space, then you still don't get the physicists definition of a vector space.
I accept that relating one trivialization to another trivialization is an imporant and interesting part of the structure of a vector bundle, but how one basis in a vector space relates to another follows from the axioms of a vector sp…

For a real $m\times n$ matrix $c_{ij}$, the following are equivalent:
(1) There exist two sets of unit vectors $\{u_i\}_{i=1}^m,\{v_j\}_{j=1}^m$ on an $(m+n)$-dimensional Euclidean space such that $c_{ij}=\langle u_i,v_j\rangle$.
(2)There are two sets of Hermitian operators $\{A_i\}_{i=1}^m$ and $\{B_j\}_{j=1}^n$ and a positive operator $\rho$ with trace $1$ in a Hilbert space $\mathcal{H}$ such that $\text{Tr }\rho =1,$ and, for every $i,j$:
$A_i B_j=B_iA_j$, $\text{spec}(A_i) \in [-1,1],$ $\text{spec}{B_j} \in [-1,1],$ and $\text{Tr}(A_i B_j W ) = c_{ij}$.