The answer to 1 is: almost yes. The "almost" is needed because of the possibility that there is an edge $e$ and a finite order element $g \in G$ such that $ge = \bar e$, in which case $g$ does not fix any vertex. The solution to that is to subdivide each edge $e$ at its midpoint, if there exists such a $g$. Having done that, one reduces to the case under the action of $G$ on $T$, no such edge $e$ exists, and now the answer is an unqualified yes. The proof is that every finite subgroup must fix a point, and therefore it must fix a vertex.
The answer to 2 is no, and it's still no for example where all edge groups are trivial. I think that you mis-applied Grushko's theorem, as I'll explain below.
For a simple counterexample, take $\Gamma$ to be the rose graph with $n$ petals, and take each $G_v$ and $G_e$ to be the trivial group. In this case $G$ is a rank $n$ free group, and the petals represent a particular free basis of $G$. But $\text{Out}(G)$ is a large and complicated group in that situation, and there are many elements which are not represented by any graph isomorphism as you suggest.
Similar counterexamples with nontrivial vertex groups are easily obtained. For example, just take the same rose graph $\Gamma$ but take $G_v$ to be your favorite finite group. In this case $\text{Out}(G)$ is still a large and complicated group that does not preserve the splitting represented by $\Gamma$.
As for Grushko's Theorem applied in the case that the edge groups are trivial, that theorem only tells you that an automorphism permutes the vertex groups, it makes no guarantee about preserving the splitting itself represented by $\Gamma$.
Added after comments: Although it remains unclear to me exactly what is the intent of your question 2, let me add a few remarks.
One thing I can think of to say is that after properly formulating an equivalence relation on the set of actions of $G$ on simplicial trees, there is a natural action of the group $\text{Out}(G)$ on the set of equivalence classes. One could equivalently describe this action with a proper formulation of an equivalence relation on the set of graph-of-groups descriptions of $G$. This action can be roughly described as in your alternative answer.
This was first done for the case of a finite rank free group $G=F_n$ and free actions on trees, in the Culler-Vogtmann paper "Moduli of graphs". Much more was done in that paper, including applications to the group $\text{Out}(F_n)$, by packaging these equivalence classes into a kind of "deformation space" for a free group (as we now think of it more generally; see below), known as the outer space of the free group.
The completely general theory was done in the Guirardel-Levitt paper "Deformation spaces of trees".
As I mentioned in the comments, free products give you actions on trees with the properties you desire.
- If $A$ or $B$ are non-free groups then the free product $A*B$ acts on its Bass-Serre tree with trivial (hence finite) edge stabilisers.
- If $A$ and $B$ are free groups then the free product $A*B$ is free but acts non-freely on its Bass-Serre tree (as vertex stabilisers are non-trivial).
I wrote out a description of the Bass-Serre tree for a free product $A*B$ and the associated action here.
More exotic examples can be cooked up using HNN-extensions or free products with amalgamation (e.g. finite but non-trivial edge stabilisers in (1)).
Best Answer
Theorem: If $G$ is a finitely generated group acting non-trivially and without inversions on a tree $X$, and if the edge stabilizers are finitely generated, then the vertex stabilizers are also finitely generated.
The theorem follows by combining three fairly elementary lemmas which I state below. The proof of these can be found in the following book:
D. E. Cohen, Combinatorial Group Theory: A Topological Approach. Cambridge etc.: Cambridge University Press (1989); ZBL0697.20001.
We say a graph of groups is finite if the underlying graph is a finite graph.
Lemma 1: If $G$ is a finitely generated group acting non-trivially and without inversions on a tree $X$, then $G$ is the fundamental group of a finite graph of groups $\mathcal{Y}$. If $G_v$ is a vertex stabilizer under the action of $G$ on $X$, then either it is isomorphic to an incident edge stabilizer or it is conjugate to a vertex group of $\mathcal{Y}$.
Lemma 2: Let $I$ be a set and $H$ an HNN extension of the form $$ H = \langle A, t_i \mid t_iB_it_i^{-1} = B_{-i} \, \forall i\in I\rangle. $$ If $H$ is finitely generated then I is finite and $A$ is finitely generated.
Lemma 3: Let $H = A \ast_C B$. If $H$ and $C$ are finitely generated then so are $A$ and $B$.