Let us consider $\mathbb{P}^1 \times \mathbb{P}^1$ and the square with vertices $(0, 0)$, $(0, 2)$, $(2, 2)$, $(2, 0)$. This corresponds to the line bundle $\mathscr{O}(2) \boxtimes \mathscr{O}(2)$. In patchworking, we were supposed to give a convex triangulation, which corresponds to a polynomial $$ \sum_{i,j=0}^2 a_{i,j} t^{v(i,j)} x^i y^j $$ with $a_{i,j} \in \mathbb{R}$, and taking a limit as $t \to 0$. Here, we had to take care of $x$, $y$ being positive and negative, so we had to actually look at a square with vertices $(\pm 2, \pm 2)$. We then associated a sign to each vertex, corresponding to the sign of $a_{i,j}$, drew the limiting real algebraic curve, and glued this to a picture on $S^1 \times S^1$.
Remark 1. We can do this with polygons with more sides. For example, if we look at the polygon with vertices $(1, 0)$, $(2, 0)$, $(2, 1)$, $(1, 2)$, $(0, 2)$, $(0, 1)$, then we are still dealing with four copies of the hexagon. If you think about how you glue these four pieces, for each edge, we need to glue two pairs. If you think about the topological structure of the moment map $X(\mathbb{R}) \to \Delta$, what we are doing is looking at the primitive integral normals of each edge, reducing this vector modulo $2$, and using this to create this pairing.