Let me use $a$ and $b$ for relative roots, so that I can later switch to $\alpha$ and $\beta$ for absolute roots.

If $b$ is a non-multipliable root, then, as you have [said](https://mathoverflow.net/questions/403345/conjugation-of-root-subgroups-by-the-weyl-group#comment1032860_403345), $V_b$ is the $b$-root space, and $X_b$ is an exponential-type map.  Specifically, it is the unique group homomorphism $V_b \to G$ whose derivative at the identity is the inclusion of the $b$-root subspace of $\operatorname{Lie}(G)$.  It can be described ‘explicitly’, for small values of explicitly, as $v \mapsto \prod X_\beta(v_\beta)$, where $\beta$ runs over the absolute roots whose restriction to $S$ is $b$, and $v = \sum v_\beta$.

If $a$ is also non-multipliable, then we have that $w_a(u)$ is the *commuting* product $\prod w_\alpha(u_\alpha)$, where $\alpha$ runs over the absolute roots whose restriction to $S$ is $a$, and $u = \sum u_\alpha$.  In particular, $w_a(u)w_a(1)^{-1}$ equals $\prod \alpha^\vee(u_\alpha)$ (or maybe the inverse of this, depending how things are normalised; I didn't check).

You have already observed that $w_a(1) X_b(w_a(1)^{-1}v)$ equals $\prod X_\beta(c_{a\beta}v_\beta)$, where $c_{a\beta} = \prod c_{\alpha\beta}$.  Now suppose that $G$ is quasi-split.  Then the set of absolute roots $\beta$ restricting to $b$ is a Galois orbit, and it is clear that $\beta \mapsto c_{a\beta}$ is constant on Galois orbits, so $w_a(1) X_b(w_a(1)^{-1}v)$ equals $X_b(c_{a b}v)$, where $c_{a b}$ is the common value of $c_{a\beta}$.  Now just conjugate by $\prod \alpha^\vee(u_\alpha)$ to finish.