On page 4, of Atiyah's $K$ theroy he stated
Suppose $V,W$ are real f.d. v.s, $E = X \times V$ and $F= X \times W$ are corresponding vector bundles. Then any homomoprhism $\varphi:E \rightarrow F$ determines a map $$ \Phi: X \rightarrow Hom(V,W), \text{ satisfying } \varphi(x,v) = (x, \Phi(x)v). $$ where $\Phi$ is continuous when $Hom(V,W)$ is considered the usual topology. Conversely, any such continuous map $\Phi:X \rightarrow Hom(V,W)$ determines a homomorphism $\varphi:E \rightarrow F$.
I would appreciate if someone can give a detailed explanation.
EDIT: Here is my attempt to spell out 3. of Bananeen's explanation.
$(\Leftarrow)$ An open set in $Hom(V,W)$, containing $\Phi(x)=(a_{i,j}(x))$, is set of matrices $A$, $|A-a_{i,j}(x)|<\varepsilon$. Continuity at each $e_j$, shows exists open sets $U_1, \ldots, U_n$ containing $x$ and that $$|(a_{i,j}(x))_{i=1}^n - (a_{i,j}(y))^{i=1}_n| < \varepsilon/2n \text{ for } y \in U_j.$$
Thus, taking $\bigcap U_i=U$, we have found an nhood of $x$, such that for all $y \in U$, $|\Phi(y)-\Phi(x)|<\varepsilon$.
$(\Rightarrow)$ The converse follows since on the basis $e_j$ vectors, $|\Phi(x)e_j - \Phi(y)e_j| \le |\Phi(x)-\Phi(y)|$. For any arbitrary vector $v$, by triangle inequality, result follows.
So to show continuity of $\Phi(x)$ from 2 we just observe that $$x \mapsto \phi(x,v) \mapsto pr_2 (\phi(x,v) ) = \Phi(x)v$$ is continuous for any $v \in V$.