Understanding Hamiltonian Path, NP vs Co-NP
A nondeterministic Turing machine is perhaps more clearly viewed as a Turing machine with two inputs: one is the actual input, in your case $\langle G,s,t \rangle$, and the other a polynomial size "witness", in your case a purported Hamiltonian path $p$. The machine checks, in polynomial time, that $p$ is indeed a Hamiltonian path from $s$ to $t$ in $G$, and if so, accepts; otherwise it rejects.

In what sense does this machine $M$ nondeterministically accept $\mathsf{HAMPATH}$?


If $\langle G,s,t \rangle \in \mathsf{HAMPATH}$ then there exists a witness $p$ such that $M(\langle G,s,t \rangle,p)$ accepts.
If $\langle G,s,t \rangle \notin \mathsf{HAMPATH}$ then $M(\langle G,s,t \rangle,p)$ rejects for all values of $p$.


Now consider the machine $!M$ which accepts when $M$ rejects and rejects when $M$ accepts.


If $\langle G,s,t \rangle \in \mathsf{!HAMPATH}$ then $!M(\langle G,s,t \rangle,p)$ accepts for all values of $p$.
If $\langle G,s,t \rangle \notin \mathsf{!HAMPATH}$ then there exists a witness $p$ such that $!M(\langle G,s,t \rangle,p)$ rejects.


As you can see, $!M$ doesn't have the same promises regarding $\mathsf{!HAMPATH}$ that $M$ has regarding $\mathsf{HAMPATH}$.