Finite Automata Input Confusion
If you're at a state $q$ reading $\alpha$ and there is no outgoing edge labeled $\alpha$, then there is way to extend the computation path. In view of that, the only valid computation path upon reading $abba$ is $q_0,q_0,q_0,q_0,q_0$; in fact, the same is true for any word ending in $a$, whose computation paths are of the form $q_0,\ldots,q_0$. Words ending in $b$ have the additional computation path $q_0,\ldots,q_0,q_1$, and words ending in $bb$ have the additional computation path $q_0,\ldots,q_0,q_1,q_2$, making three in total.