Deciding whether a 2nfa halts on every input on every branch
Lemma 1. $H_{\textsf{2nfa}}$ is decidable.

Proof.  Decide it as follows.  Given as input a two-way non-deterministic finite automaton $M_{\textsf{2nfa}}$:


Convert $M_{\textsf{2nfa}}$ into a two-way alternating finite automaton $M_{\textsf{2wafa}}$ such that the language of $M_{\textsf{2wafa}}$ is
$$L(M_{\textsf{2wafa}}) = \{x \in \Sigma^* : \text{ all execution paths of $M_{\textsf{2nfa}}$ on input $x$ halt }\},$$
that is, the set of inputs $x$ to $M_{\textsf{2nfa}}$ such that all execution paths of $M_{\textsf{2nfa}}$ on $x$ halt.  (An alternating finite automata has $\exists$-transitions and $\forall$-transitions.  Here we make every (non-deterministic) transition of $M_{\textsf{2nfa}}$ into a $\forall$-transition in $M_{\textsf{2wafa}}$, we don't use $\exists$-transitions, and we make the accept states of $M_{\textsf{2wafa}}$ be the halting states of $M_{\textsf{2nfa}}$.)
Convert $M_{\textsf{2wafa}}$ into an equivalent one-way NFA $M_{\textsf{nfa}}$. (See e.g. here.  I assume but have not verified that this conversion is computable. It appears to be computable, see the proof of Lemma 1, p. 296.)
Convert $M_{\textsf{nfa}}$ into an equivalent DFA $M_{\textsf{dfa}}$, and check whether $L(M_{\textsf{dfa}}) = \Sigma^*$ (all using standard techniques). Accept if so.  $~~\Box$