Fix the north and south poles $N, S$ of $S^2$. Given $G$, I want to compute the singular cohomology of $S^2/\sim$, where $x\sim y$ iff they lie on the same meridian between $N$ and $S$. We are then left with a space $X$ whose open sets are $\{X\}, \{X\backslash N\} = U_S,\{X\backslash S\} = U_N,$ and the open sets of $U_N\cap U_S \simeq S^1$.
To calculate the cohomology I use Mayer-Vietoris: $$ 0 \rightarrow H^0(X,G_X) \simeq G \rightarrow H^0(U_N,G_{U_N}) \oplus H^0(U_S,G_{U_S}) \simeq G \oplus G \rightarrow H^0(S^1, G_{S^1}) \simeq G \\ \rightarrow H^1(X,G_X) \rightarrow H^1(U_N,G_{U_N}) \oplus H^1(U_S,G_{U_S}) \rightarrow H^1(S^1, G_{S^1}) \simeq G \\ \rightarrow H^2(X,G_X) \rightarrow H^2(U_N,G_{U_N}) \oplus H^2(U_S,G_{U_S}) \rightarrow H^2(S^1, G_{S^1}) \simeq 0 $$
I still need to show that the middle terms are all zero, i.e. that $ \forall i>0, H^i(U_N,G_{U_N}) = H^i(U_S,G_{U_S}) = 0$.
Any hints? It doesn't seem obvious to me that $U_N$ is contractible.
NVM -- induced by the null-homotopy on Euclidean space we have a null-homotopy on $U_N$. (Check that the induced map is continuous.)