Instead of using Lie derivatives, I will stick to the basics and try to confine the proof to materials presented in the chapter itself.
First of all, there is some ambiguity in the phrasing of the the exercise itself. The author means that with the new $\tilde{g}_{\mu\nu} = \omega^2 g_{\mu\nu}$, $\xi^\mu$, as individual components, have the same numerical values as before when we had $g_{\mu\nu}$. This also means that now $\xi_\mu$, as individual components of the corresponding one-form, are numerically scaled by $\omega^2$ compared to the original $\xi_\mu$. (So the length of $\xi$ is scaled by $\omega$.)
For clarity, I will add a tilde to everything pertaining to the new metric $\tilde{g}_{\mu\nu}$. Also, I will write $\partial$ instead of $\nabla$ whenever I want to stress that it is about numerical values, rather than some tensors. Our goal is to prove the following.
$$\tilde{\nabla}_\mu \tilde{\xi}_\nu + \tilde{\nabla}_\nu \tilde{\xi}_\mu = (\tilde{\nabla}_\lambda \alpha) \tilde{\xi}^\lambda \tilde{g}_{\mu\nu}.$$
The conversions between the two versions, one with tilde and one without, are as follows.
- $\tilde{g}_{\mu\nu} = \omega^2 g_{\mu\nu}$
- $\tilde{g}^{\mu\nu} = \omega^{-2} g^{\mu\nu}$
- $\tilde{k}^\mu = k^\mu$
- $\tilde{k}_\mu = \omega^2 k_\mu$
- $\tilde{\Gamma}^\rho_{\mu\nu} = \Gamma^\rho_{\mu\nu} + \omega^{-1}(\delta^\rho_\mu \partial_\nu \omega + \delta^\rho_\nu \partial_\mu \omega - g_{\mu\nu}g^{\rho\lambda}\partial_\lambda \omega)$
Now we have $$ \begin{equation} \begin{split} && \tilde{\nabla}_\mu \tilde{\xi}_\nu + (\mu \leftrightarrow \nu) \\ &=& \partial_\mu (\tilde{\xi}_\nu) - \tilde{\Gamma}^\rho_{\mu\nu} \tilde{\xi_\rho} + (\mu \leftrightarrow \nu) \\ &=& \partial_\mu (\xi_\nu \omega^2) - \left(\Gamma^\rho_{\mu\nu} + \omega^{-1}(\delta^\rho_\mu \partial_\nu \omega + \delta^\rho_\nu \partial_\mu \omega - g_{\mu\nu}g^{\rho\lambda}\partial_\lambda \omega)\right)(\xi_\rho \omega^2) + (\mu \leftrightarrow \nu) \\ &=& \omega^2 \left(\partial_\mu \xi_\nu - \Gamma^\rho_{\mu\nu} \xi_\rho \right) + 2 \xi_\nu \omega \partial_\mu \omega - \omega(\xi_\mu \partial_\nu \omega + \xi_\nu \partial_\mu \omega - g_{\mu\nu}\xi^\lambda\partial_\lambda \omega) + (\mu \leftrightarrow \nu) \\ &=& \omega^2 \nabla_\mu \xi_\nu + \omega \left( \xi_\nu \partial_\mu \omega - \xi_\mu \partial_\nu \omega + g_{\mu\nu}\xi^\lambda\partial_\lambda \omega \right) + (\mu \leftrightarrow \nu). \end{split} \end{equation} $$
Notice that the first term, after being symmetrized with $\mu \leftrightarrow \nu$, vanishes because $\xi^\mu$ is a Killing vector with the original $g_{\mu\nu}$. Also $\xi_\nu \partial_\mu \omega - \xi_\mu \partial_\nu \omega$ is antisymmetric in $\mu$ and $\nu$ and thus vanishes as well. So we have, with $\omega^2 = e^\alpha$,
$$ \begin{equation} \tilde{\nabla}_\mu \tilde{\xi}_\nu + (\mu \leftrightarrow \nu) = 2 \omega g_{\mu\nu}\xi^\lambda\partial_\lambda \omega = \frac{2}{2} \omega^2 g_{\mu\nu}\xi^\lambda\partial_\lambda \alpha = \tilde{g}_{\mu\nu} \tilde{\xi}^\lambda \tilde{\nabla}_\lambda \alpha, \end{equation} $$
our desired result.