Beginning with
$$
F[\phi] = \int d^{4}x \ \phi^2 \ \partial_{\mu}\phi \ \partial^{\mu}\phi
$$
we replace $\phi$ with $\phi+\lambda\eta$:
$$
F[\phi+\lambda\eta] = \int d^{4}x \ (\phi+\lambda\eta)^2 \ \partial_{\mu}(\phi+\lambda\eta) \ \partial^{\mu}(\phi+\lambda\eta)
$$
Then we take the derivative w.r.t. $\lambda$ at $\lambda=0,$ and use integration by parts to remove any derivatives of $\eta$:
$$
\frac{d}{d\lambda} \left. F[\phi+\lambda\eta] \right|_{\lambda=0}
= \int d^{4}x \ \left(
2\phi\eta \ \partial_{\mu}\phi\ \partial^{\mu}\phi
+ \phi^2 \ \partial_{\mu}\eta \ \partial^{\mu}\phi
+ \phi^2 \ \partial_{\mu}\phi \ \partial^{\mu}\eta
\right) \\
= \int d^{4}x \ \left(
2\phi\eta \ \partial_{\mu}\phi\ \partial^{\mu}\phi
- \partial_{\mu}(\phi^2\ \partial^{\mu}\phi) \eta
- \partial^{\mu}(\phi^2\ \partial_{\mu}\phi) \eta
\right) \\
= \int d^{4}x \ \left(
2\phi \ \partial_{\mu}\phi\ \partial^{\mu}\phi
- \partial_{\mu}(\phi^2\ \partial^{\mu}\phi)
- \partial^{\mu}(\phi^2\ \partial_{\mu}\phi)
\right) \eta \\
$$
The $\eta$-independent factor in front of $\eta$ is now the functional derivative:
$$
\frac{\delta F}{\delta\phi}
= 2\phi \ \partial_{\mu}\phi\ \partial^{\mu}\phi
- \partial_{\mu}(\phi^2\ \partial^{\mu}\phi)
- \partial^{\mu}(\phi^2\ \partial_{\mu}\phi) \\
= 2\phi\ \partial_{\mu}\phi\ \partial^{\mu}\phi
- (2\phi\ \partial_{\mu}\phi\ \partial^{\mu}\phi + \phi^2 \partial_{\mu}\partial^{\mu}\phi)
- (2\phi\ \partial^{\mu}\phi\ \partial_{\mu}\phi + \phi^2\ \partial^{\mu}\partial_{\mu}\phi) \\
= -2\phi\ \partial^{\mu}\phi\ \partial_{\mu}\phi - \phi^2\ \partial^{\mu}\partial_{\mu}\phi
$$