Your sense of smell is off. Assuming a directed acyclic graph (the usual shape of a dependency tree), assume we write the result order to a list L. Walk over all vertices once, create a mapping M of each vertex to its number of incoming edges, and add all vertices without incoming edges to a list R. This is O(|V| + |E|). Now, pick the first item of R and append it to L. For each outgoing edge in the item we chose, decrement the associated count of its receiving vertex in M by 1. For each item where the count becomes 0, append it to R. Keep running until R is empty, and you're done. This second part of the process is again O(|V| + |E|).
We're done and found a topological order in O(|V| + |E|).
