While trying to get accustomed to the very definitions of some usual objects in geometry/topology, a lot of questions I could find complete answers to came to my mind. I will try to be clear :
Def (Vector Bundle over a base B) : Let $E$ be a manifold and $\pi: E \rightarrow B$ a submersion, $E$ is a locally trivial vector bundle of rank $r$ over B if we can cover B with open subsets : $B = \cup_{i} U_i$ such that we have local trivialisations (diffeomorphisms) : $\Phi_i : \pi^{-1}(U_i) \rightarrow U_i \times \mathbb{K}^{r}$ such that the classical diagram commutes, i.e : $ pr_1 \circ \Phi_i = \pi$.
Now we ask for a compatibility condition : on any $U_i \cap U_j$, we know from the above condition that the diffeomorphism $\Phi_i \circ \Phi_j^{-1} : U_i \cap U_j \times \mathbb{K}^r \rightarrow U_i \cap U_j \times \mathbb{K}^r$ is of the form : $$\Phi_i \circ \Phi_j^{-1}(p,v) = (p, \varphi_{ij}(p).v) $$
And we ask that $\varphi_{ij}(p)$ is a linear isomorphism instead of just a diffeo, and that it depends smoothly on p, in other words : $$\varphi_{ij}: U_i \cap U_j \rightarrow GL_r(\mathbb{K}) $$ smooth.
Now, we say that as a consequence, the vector space structure defined on a given fiber does not depend on the choice of the trivialisation (say the choice of $i$ or $j$ on the interesection).
I understand it is natural, but there is something that bugs me.
Here is how I see things : Pick a point $b$ in $U_i \cap U_j$, "at the beginning" the fiber $E_b$ is only a topological space/submanifold of E. but by identifying $E_b \rightarrow \{b\}\times \mathbb{K}^s \approx \mathbb{K}^s$ through $\Phi_i$ or $\Phi_j$ we give it a vector space structure in the following way :
For instance if $v_1,v_2 \in E_b$ we say $$ v_1 + v_2 = \Phi_i^{-1}(\Phi_i(v_1) + \Phi_i(v_2))$$ or $$ v_1 + v_2 = \Phi_j^{-1}(\Phi_j(v_1) + \Phi_j(v_2))$$ right ?
I agree this gives two different vector space structure, and we have : $$ \Phi_i(v_1) + \Phi_i(v_2) = (\Phi_i \circ \Phi_j^{-1})(\Phi_j(v_1) + \Phi_j(v_2)) $$
But : as we compose by $\Phi_i^{-1}$ or $\Phi_j^{-1}$ to go back in $E_b$ I don't see we relate the two structures on $E_b$ ? How to quantize that this structure is "independent of the choice of $i$ or $j$" ?
Given a "blank" space $V$, "how many" different vector space structures on a given field can we give to it ? Are they not isomorphic ? Does it relate to the theorem that says that if you have an homomorphism between two subsets of $R^n$ they have the same dimension (as manifolds) (so that would mean given a "blank" space, at least the dimension it will have as a vector space is unique ?
Any clarification welcome !
I think you are homing in on the right ideas, but have made some slip-ups. You say that $v_1$ is in the fiber $E_b = \pi^{-1}(b) \subseteq E$, but then you start talking about $\Phi_i(b,v_1)$ which does not make sense. You wrote that the domain of the chart $\Phi_i$ is $\pi^{-1}(U_i) \subseteq E$, but here you are applying $\Phi_i$ to $(b,v_1) \in B \times E$.
Instead, what you should be doing is saying that $\Phi_i(v_1)$ is in $\{b\} \times \mathbb{K}^r$ so that $\Phi_i(v_1) = (b,w_1)$ and, similarly, $\Phi_i(v_2)=(b,w_2)$ where $w_1,w_2 \in \mathbb{K}^r$. Then the addition of $v_1$ and $v_2$ which you define wil be based on the the operation $(b,w_1) + (b,w_2) = (b,w_1+w_2)$ in $B \times \mathbb{K}^r$.
I think after making these fixes you will have no trouble establishing that the resulting addition operation on $E_b$ is independent of which chart you used to define it.