Formally speaking, an NFA or DFA never “dies out”: for every state $q$ and for every character $a$ in the input alphabet, there is a state $q'$ such that there is a transition $q \xrightarrow{a} q'$. The state $q'$ is $\delta(q, a)$ where $\delta$ is the transition function in the definition of the automaton. The automaton only stops when it reaches the end of the input.
There is a very common convention when drawing automata to omit one hidden “dead” state $q_\dagger$ which is not accepting. Any missing transition in the drawing is a transition to $q_\dagger$, including transitions from $q_\dagger$ itself. If the automaton enters this state, it remains in this state. Nothing interesting happens once $q_\dagger$ is reached and we know that the input won't be accepted as soon as it's reached, so it's a dead state (More generally, any state from which it is impossible to reach an accepting state is called a dead state. Merging all the dead states into one doesn't change what language the automaton accepts, so it's common to speak of “the” dead state even though there could be many and some automata don't have one. If an automaton doesn't have a dead state, you can add one without changing the language; that state will be unreachable.)
So, for example, the automaton you drew has 4 states $\{q_0, q_1, q_2, q_\dagger\}$, and has 4 transitions that are not drawn: $q_0 \xrightarrow{b} q_\dagger$, $q_1 \xrightarrow{a} q_\dagger$, $q_\dagger \xrightarrow{b} q_\dagger$ and $q_\dagger \xrightarrow{a} q_\dagger$.
Hiding the implicit dead state simplifies the drawing. Making the dead state implicit, rather than defining the transition function as a partial function, simplifies reasoning about automata. It would be possible to define (non)deterministic finite automata with a partial transition function (DFAWAPTF/NFAWAPTF), and their theory would be basically identical to the theory of DFA/NFA, but reasoning would be more complicated since you'd have to have separate cases for $(q,a)$ pairs with a transition and $(q,a)$ pairs without a transition. Since every DFAWAPTF has a matching DFA and NFAWAPTF has a matching NFA, there is no point in reasoning about the WAPTF variants.
The state $q_2$ does need transitions to itself. If these transitions were not present, then an input such as $aba$ would lead to the sequence of transitions $q_0 \xrightarrow{a} q_1 \xrightarrow{b} q_2 \xrightarrow{a} q_\dagger$, so $aba$ would not be accepted. Without the transitions from $q_2$, the automaton would only accept the exact string $ab$, and not any other string starting with $ab$.