Feedback Vertex Set with vertex partitions?
Your problem reduces to Feedback Vertex Set by turning every subset $V_i$ into a clique, and creating one additional vertex $v_i$ for each subset $V_i$ that is adjacent to all vertices in $V_i$. It might be easier to think in terms of the complement problem, Maximum Induced Forest.

Since $V_i\cup \{v_i\}$ is a clique, we can select at most two vertices from this set to be in our solution (to Maximum Induced Forest). We can assume $v_i$ is always in the solution, since if less than two vertices of $V_i$ are in the solution we can safely include $v_i$, and if exactly two vertices of $V_i$ are in the solution then $v_i$ is not, but we can swap one of these two vertices (from $V_i$) with $v_i$.

A solution to Maximum Induced Forest thus selects at most one vertex from each set $V_i$ such that the result is a acyclic, and these are exactly the vertices we should select in a solution to the original problem (from the sets $V_i$ from which no vertices are selected, we can just arbitrarily select a vertex, and this vertex will be in our feedback vertex set). Conversely, it is easy to see how a solution to the original problem corresponds to a Maximum Induced Forest/Feedback Vertex Set in the graph created in this reduction.

This gives you a $O(1.7548^{n+k})$-time algorithm. If you allow parallel edges you can (instead of creating this additional vertex $v_i$) double up the edges in every clique, which has the same effect (and may give a faster algorithm, if your algorithm for solving feedback vertex set allows for parallel edges).

The problem is $NP$-hard. We can reduce from Feedback Vertex Set. To decide whether a graph has a Feedback Vertex set of size at most $s$, we create $s$ cliques of size $n+1$. We can force that one vertex of this clique is always in the solution (by including it in a singleton set $V_i$). We then create for each vertex $v$ in the original graph a set $V_i$ that contains $v$, and one vertex from each of the $s$ cliques. In at most $s$ of these sets $V_i$, we can select a vertex from each clique, but in the remaining $n-s$ sets we must select a vertex from the original graph. Therefore, we can find a solution to the original problem with feedback vertex set number $0$ if and only if the input graph had feedback vertex set number of at most $s$.