Let $A\_n$ be the dual simplicial complex to the associahedron on $n$ letters. The complex $A\_n$ is thus a simplicial triangulation of an $(n-3)$-dimensional sphere. The vertices of $A\_n$ correspond to the ways of inserting one nontrivial parentheses into the expression $a\_1 a\_2 \cdots a\_n$, and the $(n-3)$-dimensional simplices correspond to ways of inserting the maximal number ($n-2$) of parentheses into this expression. For instance, the $1$-dimensional simplices of $A\_4$ are precisely
\begin{align\*}
&((a\_1 a\_2) a\_3) a\_4\\
&(a\_1 (a\_2 a\_3)) a\_4\\
&a\_1 ((a\_2 a\_3) a\_4)\\
&a\_1 (a\_2 (a\_3 a\_4))\\
&(a\_1 a\_2) (a\_3 a\_4)
\end{align\*}
Since $A\_n$ is a triangulation of an $(n-3)$-dimensional sphere, it can be oriented (in two ways, we pick one).


**Question**: This orientation should give an orientation on each $(n-3)$-dimensional simplex $\sigma$. Such an orientation corresponds to an ordering (defined up to the action of the alternating group) of the vertices of $\sigma$, i.e., to the pairs of parentheses in $\sigma$. How can we explicitly write down this orientation?



Orienting the dual of the associahedron
There are many combinatorial models for the associahedron -- parenthesizations of $n$ variables, triangulations of the $(n+1)$-gon, planar binary trees with $n$-leaves. I'll follow the OP's lead and use parenthesizations of $n$-variables.


First, let's remember how we label the vertices of the dual associahedron. We'll label the vertices as $v\_{ij}$ for $1 \leq i < j \leq n$, $(i,j) \neq (1,n)$. The vertex $v\_{ij}$ occurs in the facet corresponding to a multiplication if the product $a\_i a\_{i+1} \cdots a\_j$ is computed as a partial step in that multiplication. For example, say we have the parenthesization
$$((a\_1 a\_2) a\_3) (a\_4 a\_5).$$
When we compute this, we compute the partial products:
$$a\_1 a\_2,\ a\_1 a\_2 a\_3,\ a\_4 a\_5$$
and hence this parenthesization will correspond to vertices $v\_{12}$, $v\_{13}$ and $v\_{45}$.


You might notice that I didn't include a vertex $v\_{1n}$, which would correspond to the entire product $a\_1 a\_2 \cdots a\_n$. Let's make a new simpicial complex $A\_n$ which does use this vertex: So we have $\binom{n}{2}$ vertices $v\_{ij}$ for $1 \leq i < j \leq n$ and, for each parenthesization, we have a facet corresponding to the $n-1$ partial products occuring when we compute that parenthesization. So $A\_n$ is a closed $n-2$ dimensional ball, with the vertex $v\_{1n}$ at the center. And the ordinary dual associahedron is the boundary $\partial A\_n$.


With that preamble, here is how to orient $A\_n$, and its boundary $\partial A\_n$. Take each multiplication and insert the multiplication symbols, so our previous example turns into $((a\_1 \ast a\_2) \ast a\_3) \ast (a\_4 \ast a\_5)$. Go across the formula from left to right, and write down the partial products in the order that they are output by that multiplication symbol. In our example, we get $a\_1 a\_2$, $a\_1 a\_2 a\_3$, $a\_1 a\_2 a\_3 a\_4 a\_5$, $a\_4 a\_5$.


**Theorem** Ordering the vertices of each facet of $A\_n$ as listed here gives an orientation of the ball $A\_n$.


To orient the boundary $\partial A\_n$, we have to work a little harder: Take a parenthesization of $n$ variables. This corresponds to a facet $\sigma$ of $A\_n$ whose vertices are ordered as above. One of those vertices is $v\_{1n}$; let's say that $v\_{1n}$ is in the $k$-th position. Delete $v\_{1n}$ from the list of vertices of $\sigma$, giving a list of vertices of $\partial \sigma$. Assign that list of vertices, in that order, the sign $(-1)^{k-1}$. In the example above, we had the list of vertices $(v\_{12}, v\_{13}, v\_{15}, v\_{45})$. The vertex $v\_{15}$ is in the $3$-rd position, so we assign the sign $(-1)^{3-1}= +1$ to the ordering $(v\_{12}, v\_{13}, v\_{45})$ in $\partial A\_5$.


**Corollary** Ordering the vertices of each facet of $\partial A\_n$ as listed here gives an orientation of the sphere $\partial A\_n$.


Why do I call this a corollary? Because this is the standard recipe for how to orient the boundary of an oriented simplicial complex. So we just need to prove the Theorem, and the Corollary will follow.


**Proof of Theorem** We just need to take two adjacent facets and check that their orientation is compatible. Two adjacent facets differ by a single association -- say we change
$${\big(} (a\_{p+1} a\_{p+2} \cdots a\_q) \ast (a\_{q+1} a\_{q+2} \cdots a\_r) {\big)} \ast (a\_{r+1} a\_{r+2} \cdots a\_s) \qquad (1)$$
to
$$(a\_{p+1} a\_{p+2} \cdots a\_q) \ast {\big(} (a\_{q+1} a\_{q+2} \cdots a\_r) \ast (a\_{r+1} a\_{r+2} \cdots a\_s) {\big)} \qquad (2).$$
All the multiplication symbols which I haven't drawn correspond to the same vertices in both $(1)$ and $(2)$. Of the two symbols which I have drawn, the left symbol gives $v\_{(p+1) r}$ in $(1)$ and $v\_{(p+1) s}$ in $(2)$; the right symbol gives $v\_{(p+1) s}$ in $(1)$ and $v\_{(q+1) r}$ in $(2)$. So our ordering has changed from
$$(\cdots, v\_{(p+1) r}, \cdots, v\_{(p+1) s},\ \cdots)$$
to
$$(\cdots, v\_{(p+1) s}, \cdots, v\_{(q+1) r},\ \cdots).$$
If we switch the second ordering to $(\cdots, v\_{(q+1) r}, \cdots, v\_{(p+1) s},\ \cdots)$, we act by a single transposition, introducing a minus sign. Thus, if we rearrange the orders to coincide on the set of vertices we have in common, then the two orders get opposite signs. This is the condition to be an orientation. $\square$


I actually found this in a very different way, but this is answer is already too long, so I'll write more in comments.


