The defining property of the inverse metric is
$$g_{\mu\nu}g^{\nu\alpha} = \delta_{\mu}^{\;\,\alpha}\tag{1}\label{eq:inverse}$$
We want to make an expansion
$$g_{\mu\nu} = \eta_{\mu\nu} + \kappa h_{\mu\nu}^{(1)} + \kappa^2 h_{\mu\nu}^{(2)}\tag{2}\label{eq:linearExp}$$
Now you can plug eq.\eqref{eq:linearExp} into the defining property \eqref{eq:inverse}, to obtain
$$\left( \eta_{\mu\nu} + \kappa h_{\mu\nu}^{(1)} + k^2 h_{\mu\nu}^{(2)}\right) g^{\nu\sigma} = \delta_{\mu}^{\;\,\sigma}$$
You can contract with the flat metric $\eta^{\alpha\mu}$ and get:
$$\begin{align}
\left( \delta^{\alpha}_{\;\,\nu} + \kappa {h^{\alpha}_{\;\,\nu}}^{(1)} + \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}\right) g^{\nu\sigma} &= \eta^{\alpha\sigma}\\
g^{\alpha\sigma} + \left(\kappa {h^{\alpha}_{\;\,\nu}}^{(1)} + \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}\right) g^{\nu\sigma} &= \eta^{\alpha\sigma}
\end{align}$$
where we have defined the perturbations with upper indices to be corresponding contractions with the flat metric. We can extract from the last line the expression:
$$g^{\alpha\sigma} = \eta^{\alpha\sigma} - \kappa {h^{\alpha}_{\;\,\nu}}^{(1)}g^{\nu\sigma} - \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}g^{\nu\sigma}\tag{3}\label{eq:selfConsistent}$$
One can then insert this equation in itself (in the right hand side) to achieve an expansion in $\kappa$ self-consistently. Let us expand up to order $\kappa^2$ like this:
$$\begin{align}
g^{\alpha\beta} &= \eta^{\alpha\beta} - \kappa {h^{\alpha}_{\;\,\nu}}^{(1)}g^{\nu\beta} - \kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)}g^{\nu\beta}\\[6pt]
&=\eta^{\alpha\beta} -\kappa \color{blue}{(\eta^{\nu\beta} - \kappa {h^{\nu}_{\;\,\lambda}}^{(1)}g^{\lambda\beta} - \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(2)}g^{\lambda\beta})} {h^{\alpha}_{\;\,\nu}}^{(1)} \\[6pt]
&\quad -\kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)} \color{blue}{(\eta^{\nu\beta} - \kappa {h^{\nu}_{\;\,\lambda}}^{(1)}g^{\lambda\beta} - \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(2)}g^{\lambda\beta})}
\end{align}
$$
where the blue terms are insertions of \eqref{eq:selfConsistent}. We can in principle repeat the process up to an arbitrary order, but we are interested on getting the expression up to order $\kappa^2$
$$\begin{align}
g^{\alpha\beta} &= \eta^{\alpha\beta} -\kappa \eta^{\nu\beta} {h^{\alpha}_{\;\,\nu}}^{(1)} + \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(1)}g^{\lambda\beta}{h^{\alpha}_{\;\,\nu}}^{(1)} -\kappa^2 {h^{\alpha}_{\;\,\nu}}^{(2)} \eta^{\nu\beta} + \mathcal{O}(\kappa^3)\\[6pt]
&= \eta^{\alpha\beta} -\kappa {h^{\alpha\beta}}^{(1)} + \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)}\color{blue}{(\eta^{\lambda\beta} - \kappa {h^{\lambda}_{\;\,\mu}}^{(1)}g^{\mu\beta} - \kappa^2 {h^{\lambda}_{\;\,\mu}}^{(2)}g^{\mu\beta})} -\kappa^2 {h^{\alpha\beta}}^{(2)} + \mathcal{O}(\kappa^3)\\[6pt]
&= \eta^{\alpha\beta} -\kappa {h^{\alpha\beta}}^{(1)} + \kappa^2 {h^{\nu}_{\;\,\lambda}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)}\eta^{\lambda\beta} -\kappa^2 {h^{\alpha\beta}}^{(2)} + \mathcal{O}(\kappa^3)\\[6pt]
&= \eta^{\alpha\beta} -\kappa {h^{\alpha\beta}}^{(1)} + \kappa^2 {h^{\nu\beta}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)} - \kappa^2 {h^{\alpha\beta}}^{(2)} + \mathcal{O}(\kappa^3)\\[6pt]
&= \eta^{\alpha\beta} - \kappa {h^{\alpha\beta}}^{(1)} + \kappa^2\left({h^{\nu\beta}}^{(1)}{h^{\alpha}_{\;\,\nu}}^{(1)} - {h^{\alpha\beta}}^{(2)}\right) + \mathcal{O}(\kappa^3)
\end{align}
$$
I hope the procedure is clear, however as you can see I don't get the factor 2 in front of the $h^{(1)}h^{(1)}$ term. Let me know if you spot mistakes.