Let us say we have a polygon $P$ in $\mathbb{R}^2$, with edges in the set $E$ (the boundary), and vertices in the set $V$. Let us say we have a point $Q$ such that $Q$ lies on one of the edges in $E$.
Let the direction of a vector not parallel to any of the edges in $E$ be denoted as $\alpha$. Consider also $\beta = \alpha+ \pi$. Note that a vector with angle $\beta$ is also not parallel to any of the edges in $E$. Given some direction angle $\gamma$, let the ray emanating from $q$ be denoted as $R_{\gamma}$
Let us consider then that direction $\alpha$ points "out of the polygon", while direction $\beta$ points "into the polygon". Then, we see that the ray $R_{\alpha}$ intersects with the $E$ once (at $q$), while the ray $R_{\beta}$ intersects with $E$ twice (once at $q$, and once more when the ray exits the polygon).
So, the discrete Jordan curve theorem cannot talk about points on the boundary of $P$?
Jordan curve theorem states that if $C\subset R^2$ is homeomorphic to $S^1$ then $R^2\setminus C$ consists of two components, one bounded, $V$, and one unbounded, $U$. Note that there is nothing here about points of $C$ since $C$ is removed. Jordan's theorem is supplemented by Schoenflies theorem which states that $V\cup C$ is homeomorphic to the closed disk.