When wikipedia says the complement, that's literally what it means:
A problem $L$ is in NP if there's a non-deterministic Turing Machine $M$ that, for every word $w \in L$, that accepts $w$ in polynomial time.
A problem $L$ is in co-NP if there's a non-deterministic TM that rejects all words $x \not \in L$ in polynomial time.
The question is, if you have the "guessing" that a non-deterministic TM gives, does it help you find yes solutions fast, or no-solutions fast?
For example, there are a bunch of $coNP$ hard problems asking, for some sets $S, T$, is $S \subseteq T$. If have an element $x \in S, x \not \in T$, then we can verify in polynomial time that the subset relation does not hold.
There is a problem in NP for every problem in coNP, and vice versa. For example, the SAT problem asks "does there exist a boolean assignment which makes this formula evaluate to True?". The complement problem, which is in coNP, asks, "do all boolean assignments make this formula evaluate to False?"
Being coNP hard just means that you can reduce in polynomial time from every other problem in coNP. So if a problem is NP-hard, its complement is coNP hard.
The reason this is interesting is, we don't actually know if $NP = coNP$. The general hypothesis is that it does not. If we assume $NP \neq coNP$, then you can prove that a problem is not NP-complete (or coNP complete) by showing that it is in both $NP$ and $coNP$.