How to prove NP-completeness of this variant of the set cover problem?
Your first guess is correct. You misinterpreted however the meaning of reductions. When we prove the hardness of a problem $A$ by a reduction from $B$, we aim to reduce $B$ to $A$ and hence, given an instance $I$ of $B$, we want to build an instance $I'$ of $A$ such that $I$ is a yes-instance if and only if $I'$ is.

Now by reducing set cover to your problem, given an instance $(U, \mathcal{F}, k)$ of set cover, where $U$ is the universe, $\mathcal{F} := \{F_1, \dots, F_m\}$ the collection of sets and $k$ the maximum number of sets in the answer. We build the instance $(U, \mathcal{F}', k')$ as follows. If $k$ equals 0, then we output No. Otherwise, if $U \in \mathcal{F}$, we output yes. Else, Let $\mathcal{F}' := \mathcal{F} \cup \{U\}$, which means we add a new set  to the collection, namely the set consisting of the whole universe. We also set $k' = k + 1$. We claim the original insatnce has a solution of size $k$ for the set cover problem if and only if the new instance has a solution of size $k+1$ of your problem.

For the first direction,  given a solution of size $k$ for thr set cover problem, add the new universal set to the solution to build a solution for your problem of size $k+1$. (Why does this work?)

Now the harder direction, given a solution for the built instance of your problem of size at most $k+1$, if the universal set is a part of thr solution, the  we can remove it to get a solution for the original insatnce of set cover with size at most $k$. If the universal set is not present, then remove any set from the solution (since you can switch this set with the universal set getting back to the previous case).

By this far we proved the hardness of the problem. For the completeness proof, note that given a solution of size $k+1$, we can count for each element in the universe how many times it was covered in the solution in polynomial time. Hence we can proof a positive witness in polynomial time and the problem is NP-complete.