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 if $\dim X = 2$ and the collapsed faces are all $2$-dimensional then 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 contractible using this notion. It should be possible to strengthen it to get "honest" collapsibility, if the definition is weakend slightly. At each step the complex may only be collapsible onto a graph; but these graphs should themselves collapse onto vertices in some later stage $X\setminus U_j$. (Perhaps I will update this answer with it on some other evening). I have not seen definitions for infinite collapsibility of this form before, perhaps they are new.

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