Confusion in Reduction of Hamiltonian-Path to Hamiltonian-Cycle
If the vertex $u$ is not added to to $G'$, then a Hamiltonian cycle in $G'$ does not necessarily correspond to a Hamiltonian path from $s$ to $t$. This is because the cycle may not have $s, t$ adjacent to each other.
For example, one could have

Since there is no Hamiltonian path from $s$ to $t$ in the first graph, there is no Hamiltonian cycle in the second graph. However, if you didn't include $u$,

then there is a Hamiltonian cycle so the reduction fails.