Let me try an argument different from Christian's: $\sigma$ does not act freely as $\chi(\mathcal O_X)=1$ and hence not divisible by $2$. At a fixed point $x$, $\sigma$ acts by $\pm1$ on the fibre of $\omega_X$ and hence acts by $1$ on the fibre of $\omega_X^{\otimes2}$. It also acts by a scalar on a global non-zero section of $\omega_X^{\otimes2}$ but as that section is non-zero at $x$ this scalar must be $1$.

Let me try an argument different from Christian's: $\sigma$ does not act freely as $\chi(\mathcal O_X)=1$ and hence not divisible by $2$. At a fixed point $x$, $\sigma$ acts by $\pm1$ on the fibre of $\omega_X$ and hence acts by $1$ on the fibre of $\omega_X^{\otimes2}$. It also acts by a scalar on a global non-zero section of $\omega_X^{\otimes2}$ but as that section is non-zero at $x$ this scalar must be $1$.

Addendum: It seems that it is essential that there are fixed points. If we look at a bielliptic example; $E\times F$ the product of two elliptic curves with $\tau$ acting by an automorphism of order $4$ on $E$ (assumed to have one) and translation by an element of order $4$ on $F$ then if we divide by $\tau^2$ we have that $\tau$ induces an involution which acts by multiplication by $-1$ on global sections of $\omega^{\otimes 2}$.