The problem is L-complete.
It’s easier to think about it when the edges are written backwards. That is, I will consider the problem formulated as follows: given a directed acyclic graph such that every node has out-degree at most $1$, and vertices $s$ and $t$, determine if $t$ is reachable from $s$.
To see that it is in L, just follow the unique path starting from $s$ until you reach either $t$, or a sink.
To see that it is L-hard, use a variant of the usual reduction showing L-hardness of undirected $s$-$t$-connectivity. Fix a deterministic logspace TM $M$; you may assume that it is endowed with a polynomially bounded clock, and whenever it is about to accept, it erases all work tapes and goes into a loop until exhausting the clock. Given an input $x$, consider the directed graph consisting of configurations of $M$, with edges given by the successor configuration relation, $s$ being the initial configuration, and $t$ being the accepting configuration (which is unique by our requirements on $M$). Since the TM is deterministic, each node has out-degree at most $1$, and since the TM is clocked, the graph is acyclic.