all unoriented bordism groups are $\mathbb{Z}/2$s

Question

Is it true that all unoriented bordism groups are $\mathbb{Z}/2$ vector spaces?

This means that any unoriented bordism groups must be of the form: $$ \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/2 \dots? $$

1. If so what is the precise proof for this? My way of seeing it is that two of $M$ bound $M\times I_1$ if we do not have orientations. Namely $$ \partial(M\times I_1) = M \sqcup M $$ So two of $M$ as M \sqcup M must be trivial null bordant. Is this precise and no loop hole rigorous?

2. However, unoriented bordism groups with spin structures such as $pin^+$ and $pin^-$ can have more than $\mathbb{Z}/2$ classes? How do we understand it? Since $$ \Omega_2^{pin^-}=\mathbb{Z}/8. $$ What is wrong with my argument in the item 1 to show that $\Omega_2^{pin^-}$ must be $\mathbb{Z}/2^k$?