Yes, Thompson's group $F$ is definably bi-orderable.
Let $a$ be some element of $F$ with the support of $a$ equal to $(0,1/2)$. Let $b$ be some element of $F$ with the support of $b$ equal to $(1/2,1)$.
We will rely upon the following facts
- If $g$ and $h$ are in $F$ then $[a^g,b^h] = 1_F$ if and only if $(1/2)g \leq (1/2)h$.
- $F$ acts transitively on the dyadic rationals.
The set $S_1:= \left\{ f \in F \mid [a,b^f] = 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f \geq 1/2$.
The set $S_2:= \left\{ f \in F \mid [a^f,b] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f > 1/2$.
The set $S_3:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ (here $d = (1/2)g$).
The set $S_4:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \text{ and } \forall h \in F \text{ we have } [a^h,b^{hf}] = 1_F \vee [a^g,b^h] = 1_F\right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ and for all dyadic rationals $e \in (0,1)$ either $(e)f \geq e$ or $e \geq d$ (here $d = (1/2)g$ and $e = (1/2)h$.
Equivalently $S_4$ is the set of non-identity elements $f$ of $F$ for which the right gradient at the infimum of the support of $f$ is strictly greater than $1$.
The union $\{1_F\} \cup S_4$ forms the cone of a bi-order on $F$. Specifically $\{1_F\} \cup S_4$ is the cone of $\preceq^+_{x^-}$ from the article of Navas and Rivas linked to in the question.
EDIT: Since people seem to be interested in the case of $[F,F]$ and Chehata's group I have added below a slightly stronger argument that applies to them.
Let $G$ satisfy the following
- Every element of $G$ has only finitely many components of support.
- For any $0 < u < v < 1$ and $0 < w < x < y < z < 1$ there is $g \in G$ with $w < (u)g < x < y < (v)g < z$.
- There is some element $a$ (that we now fix) of $G$ with a single component of support $(p,q)$ bounded away from $0$ and $1$.
$G$ could be either of $[F,F]$ and Chehata's group.
Fix a non-identity elements $b$ of $G$ with the support of $b$ a proper subset of the support of $a$.
Let $S_5$ be the set of $g$ in $G$ such that $[g,a] = 1_G = [g,b]$.
For $h \in G$ write $\bar{h}$ for the boundary of the support of $h$. If $h$ is in $S_5$ then $\mathrm{supt}(h) \cap \bar{b} = \varnothing = \bar{h} \cap \mathrm{supt}(a)$. It follows that $\mathrm{supt}(h) \cap \mathrm{supt}(a) = \varnothing$. An element of $G$ whose support is disjoint from the support of $a$ is easily in $S_5$ so $S_5$ is the set of elements of $G$ whose supports do not intersect the support of $a$.
Fix a non-identity element $c$ of $G$ with the support of $c$ a subset of $(q,1)$.
Let $S_6$ be the set of $g$ in $G$ such that there exists $h \in G$ with $[a^h,a] = 1_G = [a^h,b]$ and $[a^h,c] \neq 1_G \neq [a^h,a^g]$.
We will now argue that $S_6$ is the set of those $g$ in $G$ with $q < (q)g$.
Let $g$ be in $S_6$. There exists a conjugate $a^h$ of $a$ whose support does not intersect $\mathrm{supt}(a)$ but does intersect $\mathrm{supt}(c)$ and does intersect $\mathrm{supt}(g)$.
Any conjugate of $a$ must have a single component of support so either $\mathrm{supt}(a^h) \subseteq (0,p)$ or $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^h$ intersects the support of $c$ we must have $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^g$ intersects the support of $a^h$ it follows that the support of $a^g$ intersects $(q,1)$. Now it follows that $q < (q)g$.
If $g$ is in $G$ and $q < (q)g$ then $g$ is in $S_6$ by condition 2. above.
$S_6$ now corresponds to $S_1$ from the original proof and the rest of the construction works similarly.