For any topological group $G$, there is a classifying space $BG$ and a principal $G$-bundle $EG \to BG$ called the universal principal $G$-bundle which is determined up to isomorphism by the fact that $EG$ is weakly contractible. On a paracompact topological space $X$, any principal $G$-bundle $P \to X$ admits a map $f : X \to BG$, called a classifying map, so that $P \to X$ is isomorphic to $f^*EG \to X$. Moreover, two principal $G$-bundles $P_1, P_2 \to X$ are isomorphic if and only if their classifying maps $f_1, f_2 : X \to BG$ are homotopic. In particular, a principal $G$-bundle is trivial if and only if its classifying map is nullhomotopic.
Complex rank $n$ vector bundles can be identified with principal $U(n)$-bundles, so your problem reduces to showing that every map $f : \dot{M} \to BU(2)$ is nullhomotopic.
The long exact sequence in homotopy applied to the universal principal $G$-bundle, together with the weak contractibility of $EG$, shows that $\pi_{k+1}(BG) \cong \pi_k(G)$. In particular, $\pi_1(BU(2)) \cong \pi_0(U(2)) = 0$ as $U(2)$ is path-connected.
The surface $\dot{M}$ deformation retracts onto a bouquet of circles. Restricting $f$ to one of these circles, we get a map $S^1 \to BU(2)$ which is nullhomotopic as $BU(2)$ is simply connected. It follows that $f$ is nullhomotopic and therefore every rank two complex vector bundle on $\dot{M}$ is trivial.
As $U(n)$ is path-connected for every $n$, $BU(n)$ is always simply connected so the argument above would still work if we replace $BU(2)$ by $BU(n)$. Therefore we see that every complex vector bundle on $\dot{M}$ is trivial.