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
1. If $g$ and $h$ are in $F$ then $[a^g,b^h] = 1_F$ if and only if $(1/2)g \leq (1/2)h$.
2. $F$ 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.