A question about halt (or stop) of Turing machine
No, Turing machine are defined in a different way than finite-automata: they don't "stop" at the end of the input, they stop whenever they reach a final state $q\in F$. Usually, there are two final states, $q_{acc},q_{rej}$ – if the machine transitions to one of these final states it stops; but until then, it keeps running.

Therefore, $\delta(q,\cdot)$ must be defined for any $q\in Q\setminus F$ and and letter the head points on. This will tell us how the machine behaves when in the state $q_t$ (using your symbol, assuming it is not a final state; see below if it is).



regarding halting without accepting/rejecting: the answer is yes. If, say, $F=\{q_{acc},q_{rej}, q_{inconclusive}\}$, and the machine reached the third final state, we can say it neither accepted nor rejected, but it halts indeed since it reached a final state in $F$. Of course, $F$ can be as large as $Q$. It may make sense for computing other task than yes/no decision problems.