When does the product automaton of two NFAs A and B not decide L(A) U L(B)?
The culprit is the fact that, in the presence of nondeterminism, automata need not have transitions to other states for the same input symbols. 

Consider two NFA over the input alphabet $\{a,b\}$:

$$N_1 = (\{q_1\},\{a,b\},q_1,\delta_1,\{q_1\})$$

where $\delta_{1}(q_1,a) = \{ q_1 \}$ and $\delta_{1}(q_1,b) = \emptyset$

and

$$N_1 = (\{q_2\},\{a,b\},q_2,\delta_2,\{q_2\})$$

where $\delta_{2}(q_2,a) = \emptyset$ and $\delta_{2}(q_2,b) = \{ q_2 \}$

– that is, each of these automata has a single state, which is accepting, and a single transition, which is a loop on $a$ or $b$, respectively.

Clearly, $L(N_1)= \{ a^n \mid n \geq 0 \}$ and $L(N_2)= \{ b^n \mid n \geq 0 \}$. But the product construction will give us an automaton 

$$N_{12} = (\{ (q_1,q_2 \}, \{ a,b \}, (q_1,q_2), \delta_{12}, \{ (q_1,q_2)\})$$

where $\delta_{12}((q_1,q_2),a) = \emptyset$ and $\delta_{12}((q_1,q_2),b) = \emptyset$, since there is no symbol $x$ such that both $N_1$ and $N_2$ admit an $x$-transition from their sole state.

Therefore $L(N_{12}) = \{ \varepsilon \} \neq L(N_1) \cup L(N_2)$.