Consider the disturbed metric $\widetilde{g}_{\mu\nu}=g_{\mu\nu}+h_{\mu\nu}$ with $|h_{\mu\nu}|<1$, meaning the pertubation is small and we can use $g$ to raise and lower indices, then $\widetilde{g}^{\mu\nu}=g^{\mu\nu}-h^{\mu\nu}$ with $|h^{\mu\nu}|<1$. The sign change arises since we want to have $\widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu}=\delta_\lambda^\nu$ as well as $g_{\lambda\mu}g^{\mu\nu}=\delta_\lambda^\nu$. Neglecting the pertubation in second order, we have:
\begin{equation}
\widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu}
=(g_{\lambda\mu}+h_{\lambda\mu})(g^{\mu\nu}-h^{\mu\nu})
=\delta_\lambda^\nu
-\underbrace{g_{\lambda\mu}h^{\mu\nu}}_{=-h_\lambda^\nu}
+\underbrace{h_{\lambda\mu}g^{\mu\nu}}_{=h_\lambda^\nu}
+\mathcal{O}(h^2).
\end{equation}
Therefore we need to put $\delta g^{\mu\nu}=-h^{\mu\nu}$. Furthermore, we have:
\begin{equation}
A_{\mu\nu}B^{\mu\nu}
=A^{\kappa\lambda}g_{\kappa\mu}g_{\lambda\nu}B^{\mu\nu}
=A^{\kappa\lambda}B_{\kappa\lambda}.
\end{equation}