= Solution
For any <vector field> $Y$, apply the <Leibniz rule> to the scalar $\omega(Y)$:
$$
\mathcal L_V(\omega_aY^a)
=(\mathcal L_V\omega)_aY^a+\omega_a(\mathcal L_VY)^a.
$$
Using $\mathcal L_VY=[V,Y]$ and expanding the <Lie bracket> in a coordinate chart leaves
$$
\boxed{(\mathcal L_V\omega)_a
=V^b\nabla_b\omega_a+\omega_b\nabla_aV^b}.
$$
The connection terms cancel because the <Levi-Civita connection> is torsion-free. Applying this formula to each slot of the <metric tensor> and using <metric compatibility> gives
$$
\boxed{(\mathcal L_Vg)_{ab}=\nabla_aV_b+\nabla_bV_a}.
$$
Equivalently, one may prove both identities at a point in <normal coordinates>; since both sides are tensors, the result then holds in every coordinate system.
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