For your Minkowski background metric:
$$g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu}$$
We have that the perturbation can be written as:
$$\delta g_{\mu\nu}=g_{\mu\nu}-\eta_{\mu\nu}=\kappa h_{\mu\nu}$$
We also know that, at first order:
$$g^{\mu\nu}=\eta^{\mu\nu}-\kappa h^{\mu\nu}$$
Now we want to find it's covariant form, which goes like:
$$\delta g^{\mu\nu}=g^{\mu\lambda}\delta g_{\lambda\rho}g^{\rho\nu}$$$$\delta g^{\mu\nu}=-g^{\mu\lambda}\delta g_{\lambda\rho}g^{\rho\nu}$$
Now simply substitute into this equation from our other equations:
$$=\left(\eta^{\mu\lambda}-\kappa h^{\mu\lambda}\right)\left(\kappa h_{\lambda\rho}\right)\left(\eta^{\rho\nu}-\kappa h^{\rho\nu}\right)$$$$=-\left(\eta^{\mu\lambda}-\kappa h^{\mu\lambda}\right)\left(\kappa h_{\lambda\rho}\right)\left(\eta^{\rho\nu}-\kappa h^{\rho\nu}\right)$$
Throwing away the third order term we obtain:
$$=\kappa h^{\mu\nu}-\eta^{\mu\lambda}\kappa h_{\lambda\rho}\kappa h^{\rho\nu}-\kappa h^{\mu\lambda}\kappa h_{\lambda\rho}\eta^{\rho\nu}$$$$=-\kappa h^{\mu\nu}+\eta^{\mu\lambda}\kappa h_{\lambda\rho}\kappa h^{\rho\nu}+\kappa h^{\mu\lambda}\kappa h_{\lambda\rho}\eta^{\rho\nu}$$
$$=\kappa h^{\mu\nu}-\kappa h_{\rho}^{\mu}\kappa h^{\rho\nu}-\kappa h^{\mu\lambda}\kappa h_{\lambda}^{\nu}\eta$$$$=-\kappa h^{\mu\nu}+\kappa h_{\rho}^{\mu}\kappa h^{\rho\nu}+\kappa h^{\mu\lambda}\kappa h_{\lambda}^{\nu}\eta$$ Since the metric must be symmetric, so must the perturbation be also hence we can write:
$$\delta g^{\mu\nu}=\kappa h^{\mu\nu}-2\kappa h_{\rho}^{\mu}\kappa h^{\rho\nu}$$$$\delta g^{\mu\nu}=-\kappa h^{\mu\nu}+2\kappa h_{\rho}^{\mu}\kappa h^{\rho\nu}$$
Now I got a factor of 2 different from your reference, Which I think can be eliminated by applying requirement for the total metric:
$$g^{\mu\nu}g_{\mu\nu}=\delta_{\mu}^{\mu}$$ But I think you get the Idea, it's a process that just grows outrageously in tediousness with each higher order. Cheers!! (: