I believe there must be a name in the literature for a class of graphs I call "staged" graphs; but I cannot find it. Would anyone point me to the right direction?
An undirected graph $(V,E)$ is an $n-$stage graph in the strict sense if there exists a partition of $V$ in $n$ subsets $V_1,\dots,V_n$, such that $((V_i \times V_j) \cap E \neq \emptyset) \implies i\in\{j-1, j+1\}$.
An undirected graph $(V,E)$ is an $n-$stage graph in the loose sense if there exists a partition of $V$ in $n$ subsets $V_1,\dots,V_n$, such that $((V_i \times V_j) \cap E \neq \emptyset) \implies i\in\{j-1, j, j+1\}$.
Basically, in a stage graph vertices are partitioned into "stages", and edges only connect vertices in consecutive stages (strict definition) or vertices in consecutive stages or in the same stage (loose definition). The loose definition is the one I'm most interested in!
After posting the question, I found a related notion for directed graphs, in some sense "stricter" than both definitions above:
A directed graph $(N,A)$ is a multistage graph with $n$ stages if there exists a partition of $N$ in $n$ subsets $N_1,\dots,N_n$, such that $((N_i \times N_j) \cap A \neq \emptyset) \implies i=j-1$.
Basically, in a multistaged graph each arc must point to a node in the "next" stage - so, following an arc moves you exactly one stage "forward", it cannot move you backward or leave you on the same stage. I'd still be interested about something matching my initial definitions though, particularly the "loose" one.
A graph is an $n$-stage graph in the loose sense if and only if it has diameter at least $n-1$.
If the graph has diameter $d$ (the largest distance between any two vertices is $d$) then pick a pair of vertices $(x,y)$ such that the distance from $x$ to $y$ is $d$, and let $V_k$ be the set of all vertices at distance $k$ from $x$. The $d+1$ sets $V_0, V_1, \dots, V_d$ form a loose $(d+1)$-stage partition of the graph.
We can then combine adjacent stages in any way we like to get a loose $n$-stage partition for any $n<d+1$.
However, the graph cannot be a loose $n$-stage graph for any $n>d+1$, because in any loose $n$-stage partition, the distance between any vertex in the first stage and any vertex in the last stage is at least $n-1$.
A graph is an $n$-stage graph in the strict sense if and only if it has diameter at least $n-1$ and is bipartite.
First of all, it must be bipartite, because putting the even stages on one side and the odd stages on the other forms a bipartition. By the same argument as above, it must have diameter at least $n-1$.
Second, if we use the construction above to find a loose $(d+1)$-stage partition of a bipartite graph with diameter $d$, it will turn out to be strict. Two vertices at distance $k$ from $x$ cannot be adjacent, since otherwise we would get an odd cycle.
We cannot as easily shrink the number of stages by combining adjacent stages. But we can use a "folding-over" process: if we have a strict $k$-stage partition $V_1, V_2, \dots, V_k$, then we have a strict $(k-1)$-stage partition $$ V_1, V_2, \dots, V_{k-3}, V_{k-2} \cup V_k, V_{k-1}. $$ So once we have a $(d+1)$-stage partition, we have an $n$-stage partition for all $n<d+1$, too.