They really mean to evaluate $\hat u$ on the $g$ points (with multiplicity) in $p^{-1}(z)$, so $u(z)=[\hat u(z_1),\ldots,\hat u(z_1)]$ where $p^{-1}(z)=\lbrace z_1,\ldots,z_g\rbrace$ (with possible repetitions). The ordering doesn't matter since we passed to the quotient $\Sigma^{\times g}\to Sym^g(\Sigma)$.

They really mean to evaluate $\hat u$ on the $g$ points (with multiplicity) in $p^{-1}(z)$, so $u(z)=[\hat u(z_1),\ldots,\hat u(z_g)]$ where $p^{-1}(z)=\lbrace z_1,\ldots,z_g\rbrace$ (with possible repetitions) and the ordering doesn't matter since we passed to the quotient $\Sigma^{\times g}\to Sym^g(\Sigma)$. In particular, wherever $p$ is branched $u$ hits the big diagonal.

They really mean to evaluate $\hat u$ on the $g$ points (with multiplicity) in $p^{-1}(z)$.

They really mean to evaluate $\hat u$ on the $g$ points (with multiplicity) in $p^{-1}(z)$, so $u(z)=[\hat u(z_1),\ldots,\hat u(z_1)]$ where $p^{-1}(z)=\lbrace z_1,\ldots,z_g\rbrace$ (with possible repetitions). The ordering doesn't matter since we passed to the quotient $\Sigma^{\times g}\to Sym^g(\Sigma)$.