Create a DFA and a NFA where the word includes (two) consecutive zeroes
The difference of DFA and NFA is in the transition map (or "outgoing arrows" as you describe) and has nothing to do with "incoming arrows". Namely, if $\Sigma$ is the input alphabet then the transition function of the DFA must be uniquely defined for each state and input symbol pair

$$\delta(q,a) = p, \text { for } q,p \in Q, a\in \Sigma$$

In case of the NFA the transition function maps each pair $(q,a)$ to a subset of states $Q$
$$\delta(q,a) = Q' = \{p_1, p_2, \dots, p_n\}, \text { for } Q' \subseteq Q, a\in \Sigma$$

In simply words, for DFA for each input symbol a state must have exactly one outgoing arrow, but in the case NFA for each input symbol a state may have more than one outgoing arrows or no outgoing arrow.  

In your example $\Sigma=\{0,1\}$ and thus the DFA must have exactly two outgoing arrows for each state, but the NFA may have no outgoing arrow or up to $|Q|$ outgoing arrows.