Let $\tau$ be the group generated by 180 degree rotations of $\mathbb{R^2}$ about integer lattice points. Then, the 2-sphere $\mathbb{S^2}$ can be identified with $\mathbb{R^2}/\tau$. I don't see this identification. I was able to work out that the orbit space of $\mathbb{Z^2}$ under this action is just 4 points but I'm not sure how to use that or where to go from here.
I also thought about looking at orbit spaces of straight lines in some hope that that might give me some sort of a foliation of the orbit space $\mathbb{S^2}$ but haven't figured it out yet. Any leads are appreciated.
Note that if $R_{p},R_{q}$ are 180 degree rotations in centres $p,q$, the $R_{p} \circ R_{q}$ is the translation in vector $2(p-q)$.
Hence, every equivalence class has a representative in the square $[0,2] \times [0,2]$. Considering $R_{(1,1)}$ every equivalence class has a representative in the rectangle. Furthermore:
$[0,1] \times [0,2]$ is a fundamental domain of the action (just compute the image of $(0,1) \times (0,2)$ by any of the rotations, you will find that the image is disjoint from $[0,1] \times [0,2]$). Hence $$ \mathbb{R}^{2}/\tau \cong [0,1] \times [0,2]/\tau.$$
Note that $[0,1] \times \{0\}$ is glued to $[0,1] \times \{2\}$ by the translation $x \mapsto x + (0,2) $. This first identification gives a cylinder
The lines $\{0\} \times [0,2]$ and $\{1\} \times [0,2]$ are glued to themselves by reflection in $(0,1)$ and $(1,1)$ respectively. This "zips up" the two boundary components of the cylinder, giving a sphere as required.
Some general mathematical context: This calculation shows that there is a Euclidean orbifold, which is homeomorphic to the $2$-sphere. The orbifold points are precisely images of $\mathbb{Z}^2$. Interestingly, it shows that even though the sphere cannot be given a flat metric by the Gauss-Bonnet formula, it is possible find a flat Riemannian metric on it with some quite reasonable singularities of the metric. See Thurston's "notes on orbifolds" for more examples and general theory.