In the reduced K-theory we have the exact sequence of pair $$ \tilde{K}(S(X/A)) \to \tilde{K}(SX) \to \tilde{K}(SA) \to \tilde{K}(X/A) \to \tilde{K}(X) \to \tilde{K}(A) $$ If choose $X$ to be a surface $\Sigma$ and $A$ to be its 1-skeleton, we end up with the sequence $$ \tilde{K}(S^3) \to \tilde{K}(S\Sigma) \to \tilde{K}\left(\bigvee_{1\leq i \leq n} S^2\right) \to \tilde{K}(S^2) \to \tilde{K}(\Sigma) \to \tilde{K}\left(\bigvee_{1\leq i \leq n} S^1\right),$$ where $n$ is the number of closed curves in the 1-skeleton. Notice that I am using a cell decomposition with one 0-cell and one 2-cell and that the suspension of a bouquet of circles is a bouquet of spheres.

My problem is: since the first and the last term of that sequence are trivial, we only have to know what is the map in the middle to determine $K^{-1}(\Sigma)$ and $K(\Sigma)$, but how do I calculate this map?

I would prefer a more explicit answer.