Why does this transfer function has a second zero
Maybe this way it is simpler to understand:

$$1-1.6z^{-1}+0.8z^{-2}=\frac{z^2-1.6z+0.8}{z^2}$$

Hence,

$$\frac{2-3z^{-1}}{1-1.6z^{-1}+0.8z^{-2}}=\frac{2-3z^{-1}}{\frac{z^2-1.6z+0.8}{z^2}}=\frac{z^2(2-3z^{-1})}{z^2-1.6z+0.8}$$

whose numerator is $2z^2-3z$.