How come finding an $n/2$ size clique in a directed/ non directed graph is not in $\sf NL$? obviously the "Finding an $n/2$ size clique in a directed/ non directed graph" is not $\sf NL$ but I'd really like to understand where does it fall in terms of the definition of the $\sf NL$ definition.

I have to have an input and witness tapes, both of them can be polynomial and a logarithmic size of work stripe. So why can't I just guess an $n/2$ clique, each time put 2 vertices on the work tape, check that they share an edge, and count them. The witness tape provides me each time two vertices from left to write, in total $O({\frac{n}{2}}^2)$. In order to count that I need $O(\log ({\frac{n}{2}}^2))$ which is logarithmic. so what's lacking or wrong with this attempt?
(Turning my comments into an answer.)

You have to deterministically verify that the $\mathcal{O}(n)$ vertices on your witness tape form a clique.  Since you can only move to the right, you can only do linear amount of work for this set, however there are $\mathcal{O}(n^2)$ pairs of vertices in the clique, and you have to check every pair.  Hence you cannot check every pair and thus not verify if it is a clique or not.

Note that this is not a proof that $\mathbf{Clique}$ is not in $\mathsf{NL}$, only that this algorithm (or idea) doesn't work.