Question on Riemann's bilinear relation

Question

Source: Farkas & Kra Riemann surfaces
III.3.8 Let $P\in M$ and choose a local coordinate $z$ vanishing at $P$. We have seen that there exists on $M\setminus\{P\}$ a holomorphic differential $\tilde{\theta}$ whose singularity at $P$ is of the form $$\tilde{\theta} = {dz\over z^n},\quad n\geq 2.$$ Assume $\tilde{\theta}$ is normalized so that it has zero periods over the cycles $a_1,\ldots,a_g$. Let $$\theta = \zeta_j =\left(\sum_{l = 0}^\infty \alpha_l^{(j)}z^l\right)dz\quad \text{at }P.$$ Then, as before, $$\int_{b_j}\tilde{\theta} = {2\pi i\over n-1}\alpha_{n-2}^{(j)}.$$ Q1. So I think as before means applying the identity for two closed forms $\theta,\tilde{\theta}$ on $M$ $$\int_{\partial M}f\tilde{\theta} = \sum_{l=1}^g\left(\int_{a_l}\theta\int_{b_l}\tilde{\theta} - \int_{b_l}\theta\int_{a_l}\tilde{\theta}\right).\quad(\dagger)$$ In this case, $f = \sum_{l}a^{(j)}_l z^l$ so that $f\tilde{\theta} = \sum_{l = 0}^\infty a^{(j)}_l z^{l-n}dz$. I think the equality follows from residue theorem which I think should be $2\pi i a_{n-1}^{(j)}$. Why ${2\pi i\over n-1}a^{(j)}_{n-2}$?
Q2. The above $(\dagger)$ is for closed forms. Why can we apply this?