To show that a graph-problem is in $L$ or $NL$
The problem is NL-complete.

The fact that it's in NL, as stated in the comments, follows from the fact that this problem is a restriction of the reachability problem for directed graphs.

To see that it's NL-hard, we can reduce from the reachability problem:

Consider a graph $G=(V,E)$ and vertices $s,t$. We transform $G$ to a graph $G'$ with out degree at most $2$, as follows.

For every vertex $v$ with out-neighbors $u_1,...,u_k$ ($k\ge 3$), construct vertices $v_1,...,v_{k-1}$ where there are edges from $v_i$ to $u_i$ and to $v_{i+1}$. The incoming edges to $v$ are now directed to $v_1$, and $v$ is removed.

It is not hard to see that there is a path from $s$ to $t$ in $G$ iff there is a path from $s_1$ to $t_1$ in $G'$. Also, this reduction can be implemented in log-space. Thus, your problem is NL-complete.