I am following the 2003 notes of Andreas Gathmann for this.
A vector bundle of rank r on a scheme $ X $ over a field $ k $ is a $ k-$scheme $ F $ and a $ k-$morphism $ \pi: F \rightarrow X, $ together with the additional data consisting of an open covering $ \lbrace U_{i} \rbrace $ of $ X, $ and isomorphisms $$ \psi_{i} : \pi^{-1}(U_{i}) \rightarrow U_{i} \times \mathbb{A}^{r}_{k} $$ over $ U_{i}, $ such that the automorphism $ \psi_{i} \circ \psi_{j}^{-1} $ of $ (U_{i} \cap U_{j}) \times \mathbb{A}^{r}_{k} $ is linear in the coordinates of $ \mathbb{A}^{r}_{k} $ for all $ i,j. $
My understanding of this is that $ \pi $ looks locally like the projection morphism $ p: U_{i} \times \mathbb{A}^{r}_{k} \rightarrow U_{i} $ for open sets $ U \subset X. $
Now, we define a sheaf $$ \mathcal{G}(U) = \lbrace k-\text{morphisms}\; s:U \rightarrow F \text{ such that } \pi \circ s = \text{id}_{U} \rbrace. $$
Then the author says:
Locally, on an open subset $ U $ on which $ \pi $ is of the form $ U \times \mathbb{A}^{r}_{k} \rightarrow U, $ we obviously have $$ \mathcal(U) = \lbrace k-\text{morphisms}\; s:U \rightarrow \mathbb{A}^{r}_{k} \rbrace, $$ so sections are just given by $ r $ independent functions.
My confusion centers around the fact that $ s: U \rightarrow \mathbb{A}^{r}_{k}. $ Why is it incorrect for $ s $ to map $ U $ to $ U \times \mathbb{A}^{r}_{k}, $ since $ \pi $ maps $ U \times \mathbb{A}^{r}_{k} $ to $ U$?
Given such an $s:U\to U\times\mathbb{A}^r$, by projecting, you get a morphism $t:U\to\mathbb{A}^r$ and conversely given such a morphism $t$ you get a morphism $Id\times t$. Easy to check that these maps are inverse to each other, proving they are essentially one and the same.