I was reading Witten's paper "Fermion Path Integrals and Topological Phases" (https://arxiv.org/abs/1508.04715). He claimed that

Indeed, on an orientable 3-manifold, the eigenvalues of the Dirac operator always have even multiplicity, as we showed earlier using the antilinear operation dened in eqn. (2.60). However, on an unorientable 3-manifold, the mod 2 index can be nontrivial. A simple example is $X = \text{KB} \times S^1$, where $KB$ and the spin structure on it were defined in eqns. (3.7), (3.8), and we take the periodic spin structure on $S^1$. The only zero-mode of it is a mode that is constant in the $S^1$ direction and whose restriction to $KB$ is the zero-mode found in the last paragraph. So the mod 2 index is 1.

I sort of understand this example. But it surprises me since it is well known that mod 2 index is defined in the corresponding $K$ group while $KR^{-3}(pt)\cong KO^{-3}(pt) \cong 0$ unless the proof or result I read only apply to orientable manifolds. I want to know how to make sense of this result mathematically and in particular, how to understand the mod 2 index on nonorientable manifolds.

PS: (3.7) is the definition of Klein Bottle $KB$ which is the real plane $(x_1, x_2)$ mod out the following $$(x_1, x_2)\cong (x_1+1, x_2)\cong (x_1, x_2+1)\cong (x_1+\frac{1}{2}, x_2)$$ and (3.8) is the definition of the spin structure $$\psi(x_1, x_2)=\psi(x_1+1, x_2)=\psi(x_1, x_2+1)=\gamma_2\psi(x_1+\frac{1}{2}, -x_2)$$

I was reading Witten's paper "Fermion Path Integrals and Topological Phases" (https://arxiv.org/abs/1508.04715). He claimed that

Indeed, on an orientable 3-manifold, the eigenvalues of the Dirac operator always have even multiplicity, as we showed earlier using the antilinear operation dened in eqn. (2.60). However, on an unorientable 3-manifold, the mod 2 index can be nontrivial. A simple example is $X = \text{KB} \times S^1$, where $KB$ and the spin structure on it were defined in eqns. (3.7), (3.8), and we take the periodic spin structure on $S^1$. The only zero-mode of it is a mode that is constant in the $S^1$ direction and whose restriction to $KB$ is the zero-mode found in the last paragraph. So the mod 2 index is 1.

I sort of understand this example. But it surprises me since it is well known that mod 2 index is defined in the corresponding $K$ group while $KR^{-3}(pt)\cong KO^{-3}(pt) \cong 0$ unless the proof or result I read only apply to orientable manifolds. I want to know how to make sense of this result mathematically and in particular, how to understand the mod 2 index on nonorientable manifolds.

PS: (3.7) is the definition of Klein Bottle $KB$ which is the real plane $(x_1, x_2)$ mod out the following $$(x_1, x_2)\cong (x_1+1, x_2)\cong (x_1, x_2+1)\cong (x_1+\frac{1}{2}, x_2)$$ and (3.8) is the definition of the spin structure $$\psi(x_1, x_2)=\psi(x_1+1, x_2)=\psi(x_1, x_2+1)=\gamma_2\psi(x_1+\frac{1}{2}, -x_2)$$