It is claimed in the below proof that if we contract each $gT$ to a single vertex, this forms $X/G$. I can't see why this is true. I know you can form X/G by identifying vertices in $X$ which are in the same orbit under the action of $G$, but I can't see how these give the same constructions. Any help, direction to a result or explanation would be much appreciated.
The proof does not say (in your words) that if we contract each $gT$ to a single vertex, this forms $X/G$.
Instead, what it says is that by contraction of each tree $gT$ to a single vertex, the resulting graph (denoted $X' = X / G \cdot T$) is a tree.
Perhaps the notation $X' = X/G\cdot T$ is what's confusing you? If so, I would suggest inserting extra parentheses: $$X' = X / (G \cdot T) $$ The idea here is that $G \cdot T$ is a subgraph of the tree $X'$, and the individual connected components of $G \cdot T$ are in one-to-one correspondence with the group $G$: the map $g \mapsto gT$ is a bijection between $G$ and the set of components of the subgraph $G \cdot T$. What the quotient map $X \mapsto X / (G \cdot T) = X'$ does is to contract (or collapse) each component $gT$ to a single vertex of $X'$, and that vertex is denoted $(gT)$. As said in the last line included in your post, $X'$ is also a tree.