Calculation of Berry's phase due to monopole tunneling event of $O(3)$ NLSM on square lattice

I now realize that in doing an alternating sum over a plaquette, you have an extra $1/2$ factor which results in the following:

$\sum_{i\in plaq.}\eta_{i}\omega_i=\frac{1}{2}\eta\int dx \omega(x)=\eta\frac{1}{2}\times 4\pi=2\pi\eta$.

Therefore in evaluating the Berry phase over a plaquette, each bond cut by the discontinuity line contained in the plaquette will contribute the phase $2\pi \eta S$. Divided by four to avoid double counting, it will give us $\frac{\pi}{2} \eta S$. Therefore each bond will indeed contribute $\pi \eta S$ as explained by Haldane, since it is included in two plaquettes.