TM decidability using pigeonhole principle
I think since the input word is delimited by special symbols, which the machine cannot move past, the language accepted by such a device should be finite. We know that all finite languages are regular, and regular languages are decidable by a TM. Does it make sense for answering the question?

You can easily simulate any DFA in your model, and so your model accepts all regular languages. In particular, it also accepts some infinite languages, for example the language of all words.

Your class of machines is two-way finite automata, and it is known to accept the regular languages, that is, it coincides in power with one-way finite automata such as DFAs and NFAs.
However, that is not the answer that the setter had in mind. Rather, imagine the working of such a machine, having $m$ states, on an input word of length $n$. The head could be in $n+2$ different positions, and consequently the machine can be in $(n+2)m$ different configurations. Therefore, after that many steps, either the machine has halted (either accepting or rejecting the input), or some configuration has repeated (by the pigeonhole principle), in which case the machine is in an infinite loop, and will never accept. Hence after $(n+2)m$ steps we can already tell the fate of the input.