How come finding an $n/2$ size clique in a directed/ non directed graph is not in $\sf NL$?
(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.