$\newcommand\ldef{\mathrel{:=}}$Let $\alpha$ be the short, and $\beta$ the long, simple root of $G_2$. Thus $\langle\alpha^\vee, \beta\rangle$ equals $-3$ and $\langle\beta^\vee, \alpha\rangle$ equals $-1$. The key is that these are both odd.
Tits's elements $a_r$ satisfy $a_r^2 = r^\vee(-1)$. In place of $a_\alpha$, take $a_\alpha' \ldef \beta^\vee(-1)a_\alpha$, so that $a_\alpha^{\prime\,2} $ equals $(w_\alpha\beta^\vee + \beta^\vee)(-1)a_\alpha^2 = (\alpha^\vee + 2\beta^\vee)(-1)\alpha^\vee(-1) = 1$; and in place of $a_\beta$, take $a_\beta' \ldef \alpha^\vee(-1)a_\beta$, so that $a_\beta^{\prime\,2} = (w_\beta\alpha^\vee + \alpha^\vee)(-1)a_\beta^2 = (2\alpha^\vee + 3\beta^\vee)(-1)\beta^\vee(-1) = 1$.
We have that $(a_\alpha' a_\beta')^3$ equals $$(\beta^\vee + w_\alpha\alpha^\vee + (w_\alpha w_\beta)\beta^\vee + (w_\alpha w_\beta w_\alpha)\alpha^\vee + (w_\alpha w_\beta)^2\beta^\vee + (w_\alpha w_\beta)^2 w_\alpha\alpha^\vee)(-1)(a_\alpha a_\beta)^3 = (-2(3\alpha^\vee + 4\beta^\vee))(-1)(a_\alpha a_\beta)^3$$ and $(a_\beta' a_\alpha')^3$ equals $$(\alpha^\vee + w_\beta\beta^\vee + (w_\beta w_\alpha)\alpha^\vee + (w_\beta w_\alpha w_\beta)\beta^\vee + (w_\beta w_\alpha)^2\alpha^\vee + (w_\beta w_\alpha)^2 w_\beta\beta^\vee)(-1)(a_\beta a_\alpha)^3 = (-2(2\alpha^\vee + 5\beta^\vee))(-1)(a_\beta a_\alpha)^3,$$ so $(a_\alpha' a_\beta')^3$ equals $(a_\beta' a_\alpha')^3$.
Thus $w_\alpha \mapsto a_\alpha'$, $w_\beta \mapsto a_\beta'$ is a splitting of the map from the Tits group to the Weyl group.