From the Wikipedia definition of the Lie derivative of a tensor along a vector field, we have,
$$\mathcal{L}_X g_{\mu\nu} = X^\lambda \nabla_\lambda g_{\mu\nu} + (\nabla_\mu X^\lambda)g_{\lambda \nu} + (\nabla_\nu X^\lambda)g_{\mu\lambda}$$
Since the covariant derivative of the metric vanishes,
$$\mathcal{L}_X g_{\mu\nu} = (\nabla_\mu X^\lambda)g_{\lambda \nu} + (\nabla_\nu X^\lambda)g_{\mu\lambda}$$
But we know this must be the Killing equation, i.e. $\nabla_\mu X_\nu + \nabla_\nu X_\mu$. But it is my understanding that the covariant derivative is only acting on $X$, and not on $g$ as well, so I don't see why this is justified:
$$(\nabla_\mu X^\lambda)g_{\lambda \nu} = \nabla_\mu(X^\lambda g_{\lambda \nu}) = \nabla_\mu X_ v$$
Am I misinterpreting the definition of the Lie derivative, or is something else going on?
Evaluating $\nabla_\mu(X^\lambda g_{\lambda \nu})$ by the Leibniz rule for derivatives gives
$$(\nabla_\mu X^\lambda) g_{\lambda \nu}+X^\lambda (\nabla_\mu g_{\lambda \nu}).$$
But since the covariant derivative of the metric vanishes the second term is zero, so we do indeed have
$$(\nabla_\mu X^\lambda)g_{\lambda \nu} = \nabla_\mu(X^\lambda g_{\lambda \nu}) = \nabla_\mu X_ v.$$