Breaking cycles in network graph by adding nodes and rerouting edges
This problem is Feedback Vertex Set in disguise, and hence NP-Hard, but I'd imagine there are good heuristics out there (I don't know the references myself, maybe someone can help me out here).

More specifically, for an input graph $G = (V, E)$ with minimal FVS $S \subseteq V$, there is a solution to your problem that copies each member of $S$ once (and no other nodes are copied at all), and this is best possible.  That solution is as follows.  Define the nodes of $G'$ in the following order: first we put all the nodes in $S$ (in any order), then we put the nodes of $V \setminus S$ in topological order in the DAG $G \setminus S$, and finally we put another identical copy of the nodes of $S$ (again in any order).  The edges $(u, v) \in E$ are placed in $G'$ in the natural way: if either endpoint $u, v \notin S$ then there is only one way to place the edge in $G'$, and if both $u, v \in S$ then in $G'$ the corresponding edge goes from the first copy of $u$ to the last copy of $v$.

This graph $G'$ will have the following property: you can map the nodes of $G'$ to the nodes of $G$ in a way that bijects the edges of $G'$ with the edges of $G$.  I believe this is the right way to formalize the node-splitting process you are describing (in the node-splitting view, you would iteratively split each node $s \in S$ into the two corresponding nodes in $G'$, rerouting edges such that the incoming edges all go to one new copy $s_1$ and the outgoing edges all go to the other new copy $s_2$).

Additionally, if $G$ has $n$ nodes then $G'$ has $n + |S|$ nodes, but there is no graph $G''$ on $n + |S| - 1$ nodes with the above property.
That is because, if one can split a set of nodes $S$ to reach a DAG, then surely one also has a DAG by deleting $S$ entirely.  So the size of a minimal FVS is no more than the number of extra nodes required in $G'$.