In this answer a weak $2$-tree is a simplicial complex in which any two points can be separated by a forest. It is clear that a strong $2$-tree is also a weak $2$-tree. By passing to subdivisions, we may assume that any (specific) given subpolyhedron of a simplicial complex is a subcomplex, and we will do this implicitly when discussing trees in $2$-trees.

Define the face-edge graph $G$ of a $2$-dimensional simplicial complex
$X$ to be the following graph: its vertices are the $2$- and $1$-
dimensional faces of $X$ (so just triangles and edges). There is
an edge in $G$ between a $2$-face $\sigma$ and a $1$-face $\tau$
if $\tau\subset\sigma$.

A *free face* in a simplicial complex $X$ is a face $\tau$
which is properly contained in precisely one face $\sigma$ (which
must then be of dimension exactly $1+\dim\tau$). If $\sigma$ has
a free face $\tau$ of codimension $1$, an *elementary collapse*
can be performed which deletes both $\tau$ and $\sigma$: it is convenient
to think of this as a strong deformation retract of $X$ onto the
complement of $\sigma$. A complex which, by a sequence of elementary
collapses, can be turned into a vertex is called collapsible. If some
subdivision is collapsible the complex is called geometrically collapsible.
It makes sense to speak of collapsing a complex onto a subcomplex,
or geometrically collapsing a complex $X$ onto a subpolyhedron $Y$, and we will denote the situation that $X$ (geometrically) collapses onto $Y$ by $X \searrow Y$. (These
notions are more or less standard in PL topology / combinatorial topology.) We may subdivide our complexes as much as convenient, although we won't really need to subdivide much. I won't keep track of when a certain sequence of collapses requires / does not require subdivisions.

Given a collapse $X\searrow Y$, we can realize each elementary
collapse as a PL map in a more-or-less obvious way. Composing these
we obtain a $\pi:X\rightarrow Y$ which I will call the ``collapse
map'' below. This may not be unique -- there are probably several
reasonable ways to define the map for each elementary collapse --
but it is a strong deformation retract with the property that the
preimage of any point is a tree.

__Claim:__ Let $X$ be a weak $2$-tree. If $\sigma\in X$ is
a $2$-face, its component in $G$ is either infinite or contains
a free edge of $X$.

__Proof:__ Suppose the component of $\sigma$ in $G$ is finite
and denote by $Y$ the union of all $1$- and $2$- faces in this
component (including their vertices). Let $x,y$ be two points in
the interior of $\sigma$. Let $F\subset X$ be a forest separating
$x$ from $y$. Then $F^{\prime}=F\cap Y$ is a forest which separates
$x$ from $y$ within $Y$. If a leaf of $F^{\prime}$ is not on a
free edge, there is a path ``around'' this leaf, and it is redundant:
The leaf and its unique edge can be deleted from $F^{\prime}$, and
the resulting forest still separates $x$ from $y$ within $Y$. Deleting
all redundant leaves iteratively in this way until none remain, we end up with a nontrivial forest, and each of its leaves is in a free
edge of $Y$. ${\scriptstyle \blacksquare}$

__Claim:__ A subcomplex of a weak $2$-tree is a weak $2$-tree.

__Proof:__ Just intersect any forest which separates two given points with the subcomplex. It still intersects all paths between the points, and hence separates. ${\scriptstyle \blacksquare}$

__Claim:__ A finite weak $2$-tree collapses onto a $1$-dimensional
subcomplex.

__Proof:__ Iteratively collapse all $2$-faces with a free edge.
By the previous two claims, at each step we still have a finite weak $2$-tree,
and there is still a free edge. ${\scriptstyle \blacksquare}$

__Claim:__ A finite strong $2$-tree is collapsible.

__Proof:__ Let $X$ be a finite strong $2$-tree. Collapse it
onto a $1$-dimensional subcomplex $Y$ (a graph), and let $\pi:X\rightarrow Y$
be a corresponding PL retract satisfying that the preimage of any
point is a tree. Assume there is a nontrivial loop, and let $x,y$
be two points on this loop. Their $\pi$-preimages are disjoint trees
$T_{1},T_{2}$, which can be separated by a tree $T\subset X$. But
then $\pi\left(T\right)$ is a connected set which separates $x$
from $y$, a contradiction. ${\scriptstyle \blacksquare}$

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For lack of time to think and write in detail, I decided to include some ideas without proof. I would have preferred not to put something so "unripe" here, hopefully it is useful.

I think we can prove the following claim by induction on the number
of faces (the induction step involves examining an elementary collapse
performed in reverse).

__Claim(?):__ A collapsible $2$-dimensional complex is a strong
$2$-tree.

I will leave this for another time or for another person. I think it's easy if true (and probably true) but I did not check carefully, there could be some weird edge-case. 

It is interesting is to ask what happens in the case of an infinite
complex such as a triangulation of $\mathbb{R}^{2}$. This is probably
necessary if we want to work with quasi-isometry... I suggest calling
an infinite complex $X$ collapsible if there is an infinite chain
$U_{1}\supseteq U_{2}\supseteq U_{3}\supseteq\ldots$ of open neighborhoods
of the set of all ends of $X$ such that $X\setminus U_{i}$ is compact
and collapsible for each $i$ and $\bigcap_{i}U_{i}=\emptyset$. 

I think I have a proof that any (possibly infinite) strong $2$-tree is collapsible using this notion (perhaps I will update this answer with it on some other evening). I have not seen the definition before, perhaps it is new.

There is a decent chance these ideas extend to the higher dimensional notions.