Deciding if a topological ordering with working set of size $\le k$ exists is NP-complete, even in bipartite graphs.
We can reduce pathwidth computation to this problem. Let $G$ be a undirected graph, whose pathwidth we want to compute. Now create an directed graph $H$, with vertex $v'$ for each vertex $v$ of $G$. For each edge $(u, v)$ of $G$, create a vertex $e_{(u,v)}$ and directed edges $(v', e_{(u,v)}'), (u', e_{(u,v)}')$.
Now consider a topological ordering of $H$. Define the corresponding sequence of working sets as $X_1, \ldots, X_p$. We prove that $X_1, \ldots, X_p$ is a path-decomposition of $G$. The working sets include only vertices of type $v'$, because the other vertices do not have outgoing edges. The working sets by definition satisfy the intersection property $i \le j \le k \rightarrow X_i \cap X_k \subseteq X_j$. If there is an edge $(u, v)$ in $G$, then there are edges in $H$ from both $v'$ and $u'$ to $e_{(u,v)}'$ so there must be a working set $X_i$ with $v', u' \in X_i$. Therefore each edge is a subset of some bag of the path decomposition and therefore $X_1, \ldots, X_p$ is a path decomposition of $G$.
We can also show that any path-decomposition of $G$ corresponds to a topological ordering of $H$ (consider the elimination ordering of the underlying interval graph). $\square$
For positive results, I would suggest at looking at some analogies of pathwidth for DAGs in literature. With a quick search I found the notions of directed pathwidth and DAG-width, but seems that they are not relevant here as they are trivial for DAGs.
