Is $\mathbb{Z}^2/\langle (1,1),(1,-1) \rangle$ free over $\mathbb{Z}$? No?
Suppose $G={m_{\alpha}}$ is a basis. But $m_{\alpha}=[z_{\alpha}]$ so $(1,1)[z_{\alpha}]=[(1,1)z_{\alpha}]=[0]$. Is this correct?
To clarify: Let us do another example. Let $R=k[x,y]$ and $M=R/\langle x^2+y^2 \rangle$. Write $[r]=r+I$ where $I \subset R$ is an ideal. Now suppose $G=\{m_{\alpha}\}$ is a basis. However $m_{\alpha}=[r_{\alpha}]$ so $(x^2+y^2).[r_{\alpha}]=[0]=0_M$. So $G$ is not a basis.
But then why is $\mathbb{Z}^2/\langle (1,1) \rangle$ free?
You seem to be assuming that $\langle (1,1),(-1,1)\rangle$ refers to the ideal in the ring $\mathbb{Z}^2$ generated by the elements $(1,1)$ and $(-1,1)$. I suspect this is not the intended interpretation: rather, it is intended to just mean the $\mathbb{Z}$-submodule of $\mathbb{Z}^2$ generated by the elements $(1,1)$ and $(-1,1)$ (so the quotient will just be a $\mathbb{Z}$-module and not a $\mathbb{Z}^2$-module). With this interpretation, your argument fails immediately, since you do not know that $(1,1)[x]=[(1,1)x]$ (the quotient map is only a map of $\mathbb{Z}$-modules, not a map of $\mathbb{Z}^2$-modules). To get a correct solution, I would suggest trying to compute exactly what elements there are in the set $\langle (1,1),(-1,1)\rangle$ to explicitly determine what the cosets are.
Moreover, even if you interpret $\langle (1,1),(-1,1)\rangle$ to refer to the ideal, your argument is not correct. It shows that if there is a basis for the quotient, each element of the basis is annihilated by $(1,1)$. But this is no contradiction--it just means that the basis cannot have any elements! It is still entirely possible that the quotient is a free $\mathbb{Z}$-module, as long as its basis is the empty set.
In fact, this is exactly what happens. Since $(1,1)$ is the unit of the ring $\mathbb{Z}$, $(1,1)[x]=[x]$ for any $x$ so your argument shows not just that a basis cannot have any elements, but that any element of the quotient at all is $0$. So the quotient is just $\{0\}$, which is the free $\mathbb{Z}$-module generated by the empty set. (To be clear, this is interpreting $\langle (1,1),(-1,1)\rangle$ as the ideal--if you interpret it as a $\mathbb{Z}$-submodule, you'll get a different answer.)