 Solution

The equivariant supertrace is independent of $\beta$ because positive-energy bosonic and fermionic states pair under the supercharge. Take the short-time limit $\beta \to 0$. A finite-action path becomes constant, but the twisted boundary condition then requires $x=f(x)$; its constant saddles are exactly the fixed-point set $M^f$. Supersymmetry cancels the nonzero bosonic and fermionic fluctuations away from their zero modes. Consequently the path integral localizes to a tubular neighborhood of $M^f$, with the remaining Gaussian determinants giving the local fixed-point contribution to the equivariant index theorem.