Clarifiaction on the NP-hardness of k-SUM-variants
Yes in case of 3-sum problem they are equivalent. Let $P$ be the version where the three numbers have to be distinct and $P'$ the version where you can use the same number multiple times. The reduction $P'$ to $P$ is by solving the cases with overlapping in preprocessing (since both overlapping cases can be done in almost linear time).

The other direction is a bit more complicated. We will reduce $P$ to another variant $P''$ and from $P''$ to $P$. We start by defining $P''$: given three sets of integers $A,B,C$. Can we pick a number from each to sum to zero.

$P''$ to $P'$ is quite easy. Let $M$ be the largest number in all sets. Let $U := 3M + 1$ build the set $$S := \{a+U:a\in A\} \cup \{b + 2U: b\in B\} \cup \{c-3U:c\in C\}$$ as input to $P''$. Try to prove why the reduction is correct as an exercise.

Now the last reduction from $P$ to $P''$ is a bit more complicated. It is actually randomized. Fix a correct tripple in $S$. A random partition of $S := A\dot\cup B \dot \cup C$, we will have the numbers in the triple assigned to distinct sets with probability at least $2/3 \cdot 1/3 = 2/9$. By repeating a constant factor of $\log n$ you get a running time of $O(n\log n)$ reduction with high probability.