Why you Add Counts

When you compute dp[v] from dp[u], you are saying: every path that reaches $u$ can be extended by one edge to reach $v$. So the number of new paths to $v$ through $u$ is exactly dp[u]. Since paths through different predecessors are distinct, you add them all: dp[v] = Σ dp[u] for all edges $(u, v)$.

This is different from shortest or longest path, where you take $\min$ or $\max$. Counting problems use addition.