Rigorous proof of monotonocity of points on a transversal

Question

I am trying to rigorously prove that if we have a solution to the equation $x’ = F(x)$ and this solution crosses the transversal $\Gamma$ with parametrisation $\Gamma = a + s\vec v$ at times $t_1 < t_2 < … < t_n$, then $s_1<…<s_n$ or $s_1>…>s_n$ (where $x(t_k) = a + s_k\vec v$).
I have already proven that the solution can only cross $\Gamma$ in one direction. Formally, if we get a vector $u$ that is orthogonal to $v$ we need to prove that $f(s_k) = \tilde{Pr_u}(F(a + s_k\vec v))>0$, where $\tilde{Pr_u}=\alpha$ if $Pr_u = \alpha \vec u$. Suppose there is a point $s_{k+1}$ such that $f(s_{k+1}) < 0$. Well, $f$ is defined and continuous on $[s_k,s_{k+1}]$ which means $\exists s \in [s_k, s_{k+1}]$ such that $f(s) = 0 \implies F(a+s\vec v) \parallel \vec v$ which can’t be true since $\Gamma$ is a transversal.
I know that after that I need to apply Jordan’s theorem to $\gamma = \{x(t)| t \in [t_1,t_2]\} \cup \{a+s\vec v| s \in [s_1,s_2]\}$ and state that if the solution enters one of the components (interior or exterior) after $t_2$ then it can’t escape it. That’s where I have a problem. I know that the solution cannot cross itself since it contradicts the uniqueness of solutions. I also know that the solution cannot cross $\{a+s\vec v| s \in [s_1,s_2]\}$ because the vector field points only in one direction. I would like to prove that more formally. I thought about using the function $f$ I constructed earlier but I can’t figure out how to apply it here.
Any help will be appreciated, thanks!