First expand \begin{aligned} n(\psi , \omega)=& \exp \left(\frac{-K \sigma+\omega \psi+K \sigma \cos \psi}{D}\right) n(0 , \omega) \\ & \times\left\{1+\frac{\left(e^{-2 \pi \omega / D}-1\right) \int_{0}^{\phi} e^{(-\omega \phi-K \sigma \cos \phi) / D} d \phi}{\int_{0}^{2 \pi} e^{(-\omega \phi-K \sigma \cos \phi) / D} d \phi} \right\} \end{aligned} to fourth order in $K$ and evaluate $$\int_0^{2\pi}n(\psi,\omega)\cos(\psi)\,d\psi=\pi (K\sigma/D) n(0,\omega)\frac{ 1+(\omega/D)^2-K\sigma/D}{\left[(\omega/D)^2+1\right]^2}+{\cal O}(K^4).$$
The function $n(0,\omega)$ is determined by the normalisation $$\int_0^{2\pi}n(\psi,\omega)\,d\psi=1,$$ which gives to fourth order in $K$ the equation $$n(0,\omega)=\frac{(10-(K\sigma/D-2) K\sigma/D) (\omega/D)^2+2 (K\sigma/D+2)^2+2 (\omega/D)^4}{4 \pi \left((\omega/D)^2+1\right) \left((\omega/D)^2+4\right)}+{\cal O}(K^4).$$ We thus arrive at $$\int_0^{2\pi}n(\psi,\omega)\cos(\psi)\,d\psi=\tfrac{1}{2}(K\sigma/D)\frac{1}{1+\omega^2/D^2}-\tfrac{1}{4}(K\sigma/D)^3\frac{6+(\omega/D)^2+(\omega/D)^4}{(1+\omega^2/D^2)^3(4+\omega^2/D^2)}+{\cal O}(K^4).$$ This then gives the equation for $\sigma$, $$\sigma= \tfrac{1}{2}K \sigma \int_{-\infty}^{\infty} g\left(D \omega+\omega_{0}\right) \frac{d \omega}{\omega^{2}+1}$$ $$-\frac{K^{3} \sigma^{3}}{4 D^{2}} \int_{-\infty}^{\infty} g\left(D \omega+\omega_{0}\right)\frac{6+\omega^2+\omega^4}{(1+\omega^2)^2(4+\omega^2)}\frac{d \omega}{\omega^{2}+1}+{\cal O}(K^4).$$ The term of order $K$ agrees with the formula in the OP, the term of order $K^3$ is different, you may want to check for a typo.