I came up with a solution which was incorrect. However, one of my partial results might be interesting (or might be trivial).

First let's say the start of a $2$-bushy $T$ is the node $\sigma$.

Result: Let $v_1$ and $v_2$ be two nodes. There is a $2$-bushy tree $T_1$ starting at $v_1$ and a $2$-bushy tree $T_2$ starting at $v_2$ such that either $f$ is constant restricted to $T_1$, or $f$ is constant restricted to $T_2$, or the image of $f$ restricted to $T_1$ is disjoint from the image of $f$ restricted to $T_2$.

I thought that by applying the last leg of the trichotomy many times you could produce an injective function, but this turned out not to be the case.

To prove this, note:

Lemma: Let $T$ be a $a$-bushy tree starting at a node $v$. Let $Z$ be a closed set of $2^{ <\omega}$. Suppose that, for every $b$-bushy subtree $T'$ of $T$ starting at $v$, $f([T']) \cap Z \neq \varnothing$. Then there is an $a-b+1$-bushy subtree $T^*$ of $T$ such that $f(T^*) \subseteq Z$.

Proof of lemma: Let $T^*$ be the subtree of $T$ consisting of all vertices that are not the start of an $b$-bushy subtree whose branches are all sent to $2^{ \omega}-Z$ by $f$. Then every node of $T^*$ has at least $a+b-1$ successors in $T^*$ - if it had any fewer, it would have $b$ sucessors, each the start of a $b$-bushy subtree whose branches are sent to $2^{\omega}-Z$, hence it would be the start of a $b$-bushy subtree whose branches are sent to $2^{\omega}-Z$. By assumption, $T^*$ contains $v$. Finally $[T^*] \subseteq Z$ because any branch not in $Z$ would have an open neighborhood of branches not in $Z$, in which it would be easy to form a $b$-bushy tree starting at one of the vertices of the branch whose branches do not intersect $Z$.

Now using the lemma, take a subtree $T_1$ starting at $v_1$ and let $Z=f([T_1])$. Then either we can find a 2-bushy tree $T_2$ with $f([T_2])$ disjoint from $f([T_1])$, and we're done, or we can find a $3$-bushy tree $S_2$ with $f([S_2]) \subseteq Z$. Now pick a point in $Z$. We can either find a $2$-bushy subtree $T_2$ of $S_2$, with $f([T_2])$ not containing that point, or find a 2-bushy subtree with $f([T_2])$ constant and equal to that point. In the second case, we are done. In the first case, $f([T_2])$ is a closed subset of $f([T_1])$. Iterating, we get a 2-bushy subtree $T_3$ starting at $v_1$ with $f([T_3])$ a proper closed subset of $f([T_1])$. We can do this infinitely many times and take the limit, and $f$ applied to the limit is contained in the intersection, and then repeat the process. By transfinite induction / Zorn's Lemma, we eventually must reach a closed set containing no points, which is the empty set, a contradiction. So we're done.