Let a square be subdivided into four subsquares, and label the points by their values mod 2 under the map from $R^2$ to $Z_2$. Let the vertices of the "big square" be $a,c,g,i$. The upper left subsquare has vertices $a,b,d,e$, The upper right square has vertices $b,c,f,e$, the lower left subsquare has vertices $d,e,h,g$, and the lower right square has vertices $e,f,i,h$. There is also the "diagonal square" with vertices $b,f,h,d$. Thus the point labelled $e$ is at the center of the big square, and the vertices of the diagonal square are at the midpoints of the sides of the big square.
Now the assumptions are that the sum of the vertices of each square (recall by the labellings I mean the value in $Z_2$ assigned to the given vertex) is zero mod 2. Now consider the algebraic identity
The terms here in parentheses are $0$ mod 2 by assumption, since they are respectively the vertices of the lower left, upper right, and diagonal subsquares. And mod 2 the left side is $g$. So what we have is that $g=c$ mod 2.
Our conclusion is that, under the mapping, every square has the same value mod 2 assigned to each pair of diametricdally opposite vertices.
We can use this to show the map is constant mod 2 on the plane: Let say $k$ be the value mod 2 assigned to the origin $(0,0)$. Let $P=(x,y)$ be any point other than the origin. There is then a square having diametrically opposite vertices, one at $(0,0)$ and the other at $(x,y)$. By what we have shown above, the values mod 2 of the map at (0,0) and at $(x,y)$ must be the same, so that the map at $(x,y)$ must be $k$ as well. So the map is constant mod 2.