Proving NP-hardness of Hamiltonian Cycle problem variant
Your definition:

A relaxed-Hamiltonian cycle in $G$ is a closed walk $C$ that visits every vertex of G exactly once, except for at most one vertex that $C$ visits more than once (i.e. that vertex may repeated 2 or even more times).

I'll call this revisited vertex (for a given $C$) the nexus of $C$ and look at a case that shows that even if we know which node must be the nexus of any relaxed-Hamiltonian cycle, the problem is still NP-hard.
Consider two graphs $G_1,G_2$ for which finding a Hamiltonian cycle is NP-hard (which may be two copies of the same graph). Then we create $G$ by identifying a vertex in $G_1$ with a vertex in $G_2$, with the single vertex thus created called $w$. $w$ is an articulation vertex so $G$ cannot have a (standard) Hamiltonian cycle.

If a relaxed Hamiltonian cycle exists in $G$, it must have $w$ as its nexus. To visit all nodes in $G$, we have to visit $w$ and that is the only node which allows us to cross into $G_2$ and visit all nodes there. So we have the second (minimum) visit and $w$ is the nexus.
The definition of the relaxed Hamiltonian cycle means that we could in principle find disjoint cycles all including $w$ within either $G_1$ or $G_2$ and still meet the relaxed-Hamiltonian standard. We can avoid this case by ensuring that nodes in $G_1$ and $G_2$ that we pick for identification to create $w$ are maximum degree $3$, so we cannot re-enter a second cycle in the same graph part from $w$.
Then the problem of finding a relaxed Hamiltonian cycle in $G$ is as hard as finding Hamiltonian cycles in both $G_1$ and $G_2$ which we know can be NP-hard.
Further to your claim, we therefore know that if we find a relaxed Hamiltonian cycle for $G$ we have found a Hamiltonian cycle for both $G_1$ and $G_2$ (and $G_2$ is allowed to be a copy of $G_1$).