Let $G$ is a simple undirected graph. Suppose $G$ has two subgraphs $G_1$ and $G_2$, such that $E(G_1)\cap E(G_2) =\emptyset$ ($E(G_i)$, stand for the set of edges of $G_i$). Then is it true that genus of $G$ is greater than or equal to the sum of genera of $G_1$ and $G_2$?
No. The two subgraphs can share the surface more efficiently than that. Take a graph $G$ with genus $g\ge 1$ and duplicate each edge. If you don't like double edges, subdivide them with new vertices. Then you can divide the new graph into two edge-disjoint subgraphs homeomorphic to $G$, therefore each having genus $g$, yet you can still draw the whole graph on the same surface.