How many disjoint edge-cuts a DAG must have?
Short answer: no.

Let $G$ be a complete DAG (transitive tournament) on $n$ vertices with $s$ and $t$ its source and sink, and let $k=\sqrt{n/3}$. Observe that there can be at most four disjoint cuts that contain more tham $n/3$ edges incident to $s$ or more than $n/3$ edges incident to $t$. So, if there are to be many disjoint cuts, we can assume that there exists a cut $C$ that does not contain large numbers of edges incident to $s$ and $t$.

Now let $X$ be the complete subgraph induced in $G$ by the set of vertices $x$ such that edges $sx$ and $xt$ do not belong to $C$. The number of vertices in $X$ is at least $n/3$, because otherwise $C$ would touch too many edges incident to $s$ or $t$. However, $X\setminus C$ cannot contain a $k$-path, because if such a path existed it could be concatenated with $s$ and $t$ to form a long path in $G\setminus C$. Therefore, the longest-path layering of $X\setminus C$ has fewer than $k$ layers, and has a layer containing more than $(n/3)/k=k$ vertices. Since this is a layer of the longest path layering, it is independent in $X\setminus C$, and therefore complete in $C$, so $C$ contains a path $P$ through the vertices of this layer, of length $k$. This path must be disjoint from all of the other cuts.

Every cut that is not $C$ must contain either the edge from $s$ to the start of path $P$ or the edge from the end of path $P$ to $t$, or else it would not block the path $s$–$P$–$t$. So if $C$ exists, there can be at most three disjoint cuts. And if $C$ does not exist (that is, if all cuts cover more than $n/3$ edges incident to $s$ or to $t$) there can be at most four disjoint cuts. Either way, this is a lot fewer than $k$ cuts.