Q:

Transitions Maps on a Vector Bundle are Linear

Let $E$ be a vector bundle, with $X$ base space and $p:E\to X$ a surjective projection.
Let $x\in X$ be given. Let $U_1, U_2$ be two neighborhoods of $x$ in $X$ which carry local trivializations, that is, $$\varphi_1:p^{-1}(U_1)\cong U_1 \times V_1$$and $$ \varphi_2:p^{-1}(U_2)\cong U_2 \times V_2$$ are two homeomorphisms, where $V_i\in obj(Vect_{\mathbb{C}})$ for $i=1,2$.
Then clearly, $$\left.\varphi_1\right|_{U_1\cap U_2}:p^{-1}(U_1)\cap p^{-1}(U_2)\cong U_1\cap U_2 \times V_1$$ and $$\left.\varphi_2\right|_{U_1\cap U_2}:p^{-1}(U_1)\cap p^{-1}(U_2)\cong U_1\cap U_2 \times V_2$$ are two homeomorphisms and thus $$\left.\varphi_2\right|_{U_1\cap U_2}\circ \left(\left.\varphi_1\right|_{U_1\cap U_2}\right)^{-1}: U_1\cap U_2 \times V_1 \cong U_1\cap U_2 \times V_2 $$ is also a homeomorphism. 
If $p_2:U_1\cap U_2\times V_2 \to V_2$ is the projection onto the second component, define the map $t_x:V_1\to V_2$ by $$ V_1\ni v_1 \mapsto p_2(\left.\varphi_2\right|_{U_1\cap U_2}\circ \left(\left.\varphi_1\right|_{U_1\cap U_2}\right)^{-1}((x,v_1))) \in V_2 $$ 
My question is: how do you see that $t_x$ is a linear homeomorphism and not merely a homeomorphism?

A:

Note that $\varphi_i : p^{-1} (U_i) \to U_i \times V_i$ are not mere homeomorphisms: they are local trivialization. That is, the following diagram commute
$\require{AMScd}$
\begin{CD}
    p^{-1}(U_i) @>\varphi_i>> U_i \times V_i\\
    @V p V V\ @VV p_i V\\
    U_i @>>id> U_i
    \end{CD}
and $\phi_i$ is linear when restricted to each fiber. This latter fact implies that the transition map is linear. 

