$\require{AMScd}$
As I understand it, a rank $k$ vector bundle is a pair of topological spaces with a map between them
$$
E\xrightarrow{p}B
$$
such that there exists an open cover $(U_\alpha)$ of $B$ over which $E$ locally looks like $U_\alpha\times\mathbb R^k$. Precisely, we have homeomorphisms
$$
\phi_\alpha\colon p^{-1}(U_\alpha)\to U_\alpha\times\mathbb R^k
$$
such that the diagram
\begin{CD}
p^{-1}(U_\alpha) @>\phi_\alpha>> U_\alpha\times\mathbb R^k\\
@VVbV @VV\text{pr}_1V\\
U_\alpha @>\text{id}>> U_\alpha
\end{CD}
commutes (I can't do diagonal arrows).
A section of the vector bundle is a continuous map $s\colon B\to E$ such that $p\circ s=\text{id}_B$. A basic fact about vector bundles is that every vector bundle admits at least one section, namely the zero section, obtained by identifying each $p^{-1}(U_\alpha)$ with $U_\alpha\times\mathbb R^k$ and taking $s(b)=0$. Explicitly, we are taking:
$$
s(b)=\phi_\alpha^{-1}((b,0))
$$
for $b\in U_\alpha$.
But in order to get a well defined continuous map, we need to ensure that $\phi_\alpha^{-1}(b,0)=\phi_\beta^{-1}(b,0)$ if $b\in U_\alpha\cap U_\beta$. Now one thing we certainly can't do is require that $\phi_\alpha$ and $\phi_\beta$ actually agree on the intersection – then we could glue them all together to get a global trivialization of the bundle, which we can't do in general. Why then are we able to insist that they agree at a certain point? To me, it seems equivalent to the existence of a global section, and hence completely circular.
There must be something obvious I'm missing. In the meantime, here's an equivalent formulation of the problem: we define an affine bundle $E\xrightarrow{p}B$ to be the same as a vector bundle, but we identify the fibres with $k$-dimensional affine linear subspaces of some fixed real vector spaces. So they needn't contain a special zero point.
Since affine spaces are homeomorphic to vector spaces of the same dimension, any rank $k$ affine bundle is isomorphic to some rank $k$ vector bundle (though I'm not at all sure about whether this is true, for similar reasons). So an affine bundle should admit a global section, but there is no easy way to see how to get one.
Best Answer
In your definition you are missing the condition that the transition maps for a vector bundle are supposed to be linear fiberwise. Once you assume that, the problem you're facing in gluing will go away.