The Multicut problem asks for a minimum cut separating certain pairs of vertices: formally, given a graph $G$ and demand graph $H$ on a set $T\subseteq V(G)$ of terminals, the task is to find a minimum-weight set $C$ of edges of $G$ such that whenever two vertices of $T$ are adjacent in $H$, they are in different components of $G\setminus C$. Colin de Verdière [Algorithmica, 2017] showed that Multicut with $t$ terminals on a graph $G$ of genus $g$ can be solved in time $f(t,g)n^{O(\sqrt{g^2+gt+t})}$. Cohen-Addad et al. [JACM, 2021] proved a matching lower bound showing that the exponent of $n$ is essentially best possible (for every fixed value of $t$ and $g$), even in the special case of Multiway Cut, where the demand graph $H$ is a complete graph. However, this lower bound tells us nothing about other special cases of Multicut such as Group 3-Terminal Cut (where three groups of terminals need to be separated from each other). We show that if the demand pattern is, in some sense, close to being a complete bipartite graph, then Multicut can be solved faster than $f(t,g)n^{O(\sqrt{g^2+gt+t})}$, and furthermore this is the only property that allows such an improvement. Formally, for a class $\mathcal{H}$ of graphs, Multicut$(\mathcal{H})$ is the special case where the demand graph $H$ is in $\mathcal{H}$. For every fixed class $\mathcal{H}$ (satisfying some mild closure property), fixed $g$, and fixed $t$, our main result gives tight upper and lower bounds on the exponent of $n$ in algorithms solving Multicut$(\mathcal{H})$.