Let $N\leq M \leq G$. If $N \trianglelefteq M$ and $M \trianglelefteq G$, do we have $N \trianglelefteq G$ Let $N\leq M \leq G$. If $N \trianglelefteq M$ and $M \trianglelefteq G$, do we have $N \trianglelefteq G$ ? I claim that it is not true. Because there may exists some elements in $G$ such that it does not commute with elements in $N$. But I do not know how to contruct a counterexample. 
 A: Of course in abelian groups all subgroups are normal. Try the smallest non-abelian group. It is the symmetric group on three letters, of order 6, but there is only one non-trivial, proper, normal subgroup, so no example here. Now you should have seen a certain non-abelian group of order 12. There is an example to be found there.
A: Choose $G = D_8$ with $K = \langle sr^3, r^2\rangle$ and $H = \langle sr^3\rangle$. Observe that $H \leq K \leq G$ and $H \trianglelefteq K$ and $K \trianglelefteq G$ but $H \not\trianglelefteq G$.
To address the HOW you construct the counterexample: you should have a few groups at your disposal, like the symmetric group, dihedral group, Klein 4-group, $n\mathbb Z$, the quaternion group and a few others.
It comes down to just playing around with the ones you know that aren't trivial, like abelian groups are out, the quaternion group is out because each group is normal in the quaternion group. I chose the dihedral group and just picked off the normal groups, and looked at the groups that were normal in those. Once I found some I checked if those were normal in the complete group.
Drawing a lattice diagram helps too!
