"Show that $\chi (G)\leq k$ if and only if $G$ admits an orientation such that the longest path has length at most $k-1$."
The greedy algorithm may be helpful for proving one direction of the equivalency. I would like to adapt Brooks' theorem, but so far I have not come up with anything relevant.
On
Suppose $D=(V,E)$ is a directed graph containing no (simple) directed path of length $k$; I will show that $\chi(D)\le k$.
Let $D_0=(V,E_0)$ be a maximal acyclic spanning subgraph of $D$.
Define $f:V\to\{0,\dots,k-1\}$ by letting $f(v)$ be the maximum length of a directed path in $D_0$ with initial vertex $v$. I claim that $f$ is a proper coloring of $D$, i.e., that $uv\in E\implies f(u)\ne f(v)$.
First note that, if $uv\in E_0$, then $f(u)\gt f(v)$, since $D_0$ is acyclic.
Now suppose $uv\in E\setminus E_0$. Because of the maximality of $D_0$, there is a path from $v$ to $u$ in $D_0$, say $v=v_0,v_1,v_2,\dots,v_n=u$. Then we have $f(v)=f(v_0)\gt f(v_1)\gt f(v_2)\gt\cdots\gt f(v_n)=f(u)$, i.e., $f(u)\lt f(v)$.
Both proofs can be done using a greedy algorithm.
Given a $k$ coloring of $G$, number the colors $1,\dots,k$. Take any edge adjacent to a vertex colored 1, and orient it away from that vertex. Repeat with each color, each time orienting all unoriented edges away from the current color being considered.
Given such an orientation, color any vertex without a neighbor with color 1 (such a vertex must exist, otherwise the graph would have a cycle, or path of length $\infty$). Repeat $k-1$ times, each time coloring all vertices without an uncolored neighbor.