Does it follow that each vertex stabiliser is finitely presented: No.

Just notice that there exist infinitely presented groups $H$ with an automorphism such that the corresponding semidirect product $G=H\rtimes\mathbf{Z}$ is finitely presented. This is a non-trivial graph of groups decomposition, namely an HNN-decomposition where the vertex group and edge group coincide; in case you find this too degenerate, just add some further flourish to the graph of groups, so that $H$ still appears as a vertex group and the resulting group is still finitely presented. One example is when $G=BS(1,2)^2=\mathbf{Z}[1/2]^2\rtimes\mathbf{Z}^2$ where the generators of $\mathbf{Z}$ act by the diagonal matrices $(2,1)$ and $(1,2)$, and $H=\mathbf{Z}[1/2]^2\rtimes\mathbf{Z}$ where $\mathbf{Z}$ acts by the diagonal matrix $(2,1/2)$.

On the other hand, for the second question "How about if every edge stabiliser is finite?", the answer is yes. Indeed, write $H$ as a limit of truncated presentations $H_n$ (this limit here means: inductive limit of surjections between f.g. groups). For $n$ large enough (say $n\ge n_0$), all finite subgroups of $G$ occurring in the graph of groups decomposition of $G$ are well-defined, and this allows to write $G$ as (non-attained) limit of corresponding graph of groups $G_n$ for $n\ge 0$, where in the graph of groups, $H$ was replaced with $H_n$). The argument still works if the edge groups are finitely presented instead of finite.