Confusion regarding the Definition of 2 Way Finite State Automata? Here is a description I read from one of the sources about 2WNFA:

"Unlike ordinary finite automata, a 2DFA needs only a single accept state
and a single reject state. We can think of it as halting immediately when
it enters one of these two states, although formally it keeps running but
remains in the accept or reject state. The machine need not read the entire
input before accepting or rejecting. Indeed, it need not ever accept or reject at all, but may loop infinitely without ever entering its accept or reject state."

I am unclear what it means by: "..The machine need not read the entire
input before accepting." 

For example in the automata M = ({qo, ql, q2, q3, q4, q5}, {1}, δ, qo, {q5})
(δ is transition function, L=Left, R=Right) and the transitions being:

δ(q0, 1) = (q1, R);
δ(q1, 1) = (q2, R); 
δ(q2, 1) = (q3, L); 
δ(q3, 1) = (q4, L);
δ(q4, 1) = (q5);

Does the above 2WNFA, accept any string (example: "1") as the first 4 transitions can be ignored as they bring the machine back to the 1st character in the input string, thus for the last transition to the final accepting state a single 1 is enough? On other hand I think the language it accepts is empty? 

Terribly confused because of the bidirection along with above explanation.
Unlike ordinary finite automata, a 2DFA needs only a single accept state and a single reject state.


Finite automata do not need more accepting states. Well, non-deterministic FA don't; just add two final states and appropriate $\varepsilon$-transitions. Deterministic FA, though, need multiple final states; above construction doesn't work, and Myhill-Nerode tells us that we indeed get multiple final states for some languages.


  Unlike ordinary finite automata, a 2DFA needs only a single accept state and a single reject state.


If you add rejecting states to finite automata (an easy enough exercise), neither do finite automata. Consider, for instance,

$\qquad L_0 = \{ w \mid w_1 = 0 \} \cap L$ 

for any regular $L$. Then, a finite automaton for $L_0$ can reject immediately if the first symbol is not a $0$.

In summary, I think the authors try to read meaning into arbitrary syntactical differences.

That said, note how 2DFA need a rejection state for the stated reason: there is no natural "end of computation".