I understand the following: one can classify rank 2 vector bundles over $S^2$ by considering two open sets, say slightly extending northern and southern hemispheres. These are contractible, cover $S^2$, and their intersection is $S^1$. Any vector bundle is trivial when restricted to the open sets, so we care only about gluing two trivial bundles on $S^1$, with structure group $GL_2(\mathbb{R})$, i.e. we care about functions $f:S^1\to GL_2(\mathbb{R})$. I also understand that homotopic maps will induce the same bundle, so what we really need is $\pi_1(GL_2(\mathbb{R}))$, which I'm told is $\mathbb{Z}$.
My question is, does the group structure of $\mathbb{Z}$ mean anything in terms of the vector bundles?
It is of course yes.
For example, there is a distinguished element $0\in \mathbb{Z}$, which corresponds to trivial bundle over $S^2$. Similarly the negation $n\mapsto -n$ on $\mathbb{Z}$ corresponds to reversing the orientation of the vector bundle.
The addition is slightly more complicated to give a simple geometric meaning, but the point is that the Euler class gives an explicit bijection between the (K-theoretic) set of rank 2 vector bundles over $S^2$ and $\mathbb{Z}$.
Edit: By the by, I didn't really explain why the Euler class $\in H^2(S^2)\cong \mathbb{Z}$ can be identified with the same number $\in \pi_1(GL_2(\mathbb{R}))(\cong \pi_1(\mathbb{C}^\times)\cong \mathbb{Z})$. For this, the best way is to introduce the first Chern class $c_1$(which is equal to the Euler class) and appeal to the fact that the first Chern class map is an isomorphism from the "group" of complex line bundles (=oriented real vector bundles of rank 2) over $S^2$ with the tensor product to $H^2(S^2)$, as is the clutching function construction.