Specific example of a problem that shows why **NP** isn't low for itself
As Pål GD mentions in the comments, $\mathsf{NP}^\mathsf{NP}$ is usually knows as $\mathsf{\Sigma}_2^P$, the second level of the polynomial hierarchy. If $L$ is any $\mathsf{\Sigma}_2^P$ complete problem, then $\mathsf{NP} \neq \mathsf{\Sigma}_2^P$ is equivalent to $L \in \mathsf{\Sigma}_2^P \setminus \mathsf{NP}$. So any $\mathsf{\Sigma}_2^P$-complete problem is a concrete example for a problem in $\mathsf{\Sigma}_2^P \setminus \mathsf{NP}$ under the reasonable assumption that $\mathsf{\Sigma}_2^P \setminus \mathsf{NP} \neq \emptyset$. In other words, these are the problems "most likely" to be the concrete examples you seek.

One canonical complete problem for $\mathsf{\Sigma}_2^P$ is $\exists \forall\mathsf{SAT}$: given a formula $\varphi$ on variables $\vec{x},\vec{y}$, is is true that there exists an assignment $\alpha$ to $\vec{x}$ such that $\varphi|_{\vec{x} = \alpha}$ becomes a tautology? It is easy to see that this can be solved by an $\mathsf{NP}^\mathsf{NP}$ machine: the machine guesses $\alpha$, and uses the oracle to verify that $(\lnot \varphi)|_{\vec{x} = \alpha}$ is unsatisfiable.

Generally speaking, the assumption $\mathsf{NP} \neq \mathsf{\Sigma}_2^P$ seems stronger than $\mathsf{P} \neq \mathsf{NP}$, and when proving conditional results, we prefer assuming the latter rather than the former. But it is generally conjectured that the entire polynomial hierarchy is strict, a much stronger assumption.