The Lie derivative of any tensor field $K$ with respect to a vector field $X$ \xi$ is by definition $$L_X(K)=-\frac $L_\xi(K)=-\frac d{dt}|_{t=0}(\phi_t)_*K,$$ where $\phi_t$ is the local flow of $X$. \xi$. Now, if $M$ has a Riemannian metric $g$ and $X$ \xi$ is Killing with respect to $g$, each $\phi_t$ is a local isometry of $(M,g)$. From the uniqueness of the Levi-Civita connection, it follows that every isometry $\phi$ is affine, i.e. $$\phi_*(\nabla_XY)=\nabla_{\phi_*X}\phi_*Y.$$ $\phi_*(\nabla_X Y)=\nabla_{\phi_*X}\phi_*Y.$$ From here you get immediately $\phi_*R=R$ for the Riemannian curvature, and since the Ricci tensor of $R$ is obtained by a trace: $$Ric(X,Y)=trace(V\mapsto R_{V,X}Y),$$
one gets $\phi_*Ric=Ric$ for every isometry $\phi$. The first formula thus shows that $L_X L_\xi Ric=0$ for every Killing vector field $X$.\xi$.
Edit: Here is another, purely tensorial, proof of the same statement. Let $\xi$ be Killing, in the sense that $g(\nabla_X\xi,Y)+g(X,\nabla_Y\xi)=0$ for all vector fields $X,Y$. After taking the covariant derivative wrt some vector field $Z$, and doing some standard manipulations, one gets the usual Kostant formula: $$\nabla^2_{X,Y}\xi=R_{\xi,X}Y,\qquad\forall X,Y\in C^\infty(TM).$$ This is just a rewriting of $$L_\xi(\nabla_XY)=\nabla_{L_\xi X}Y+\nabla_X(L_\xi Y),$$ i.e. some sort of Leibniz formula. Applying this formula several times eventually yields the corresponding Leibniz formula for $R$: $$L_\xi(R_{X,Y}Z)=R_{L_\xi X,Y}Z+R_{X,L_\xi Y}Z+R_{X,Y}(L_\xi Z),$$ i.e. $L_\xi R=0$, and finally $L_\xi Ric=0$ after taking the trace.
Of course, this is just the infinitesimal version of the first proof...
