The comments section was getting unwieldy, so I'll answer here. Hopefully this is helpful.

What I was trying to say is as follows: suppose that $(M,g,X)$ is a steady soliton, i.e. $$ \mathcal{L}_X(g) = 2 Ric_g. $$ Then, let $\Phi_t$ denote the flow of $-X$. You may check that $h(t) :=\Phi^*_tg$ is a solution to the Ricci flow. Then, the Ricci curvature of $h(t)$ is $$ Ric_{h(t)} = Ric_{\Phi_t^*g} = \Phi_t^*Ric_g. $$ Thus, $$ \frac{d}{dt} Ric_{h(t)} = -\mathcal{L}_X Ric_g $$ Now, relating $\mathcal{L}_X Ric_g$ to $D_X Ric_g$ in the usual way: \begin{align*} \mathcal{L}_X Ric(V,W) & = X(Ric(V,W))-Ric([X,V],W)-Ric(V,[X,W])\\ & = D_X Ric(V,W) -Ric(D_XV+[X,V],W)-Ric(V,D_X W +[X,W])\\ & = D_X Ric(V,W) -Ric(D_VX,W)-Ric(V,D_WX) \end{align*} and using the formula for $\frac{d}{dt} Ric_{h(t)}$ under the Ricci flow, you may find an equation for $D_X Ric$ (for example, it is the equation found in the paper of Brendle's I linked above; I've matched his sign conventions with $X$).

By the way, you can simplify the above argument by using the Uhlenbeck trick (if you work through the details of what I did above, you'll see that there is a good deal of cancellation, which you can exploit by using $D_{\frac{d}{dt}}$ instead of the time derivative above. See, e.g. ch 6 of this book: http://download.springer.com/static/pdf/28/bok%253A978-3-642-16286-2.pdf?auth66=1385658655_8c41399e85b51a3addd04347ad203f0d&ext=.pdf among many other places.

The comments section was getting unwieldy, so I'll answer here. Hopefully this is helpful.

What I was trying to say is as follows: suppose that $(M,g,X)$ is a steady gradient soliton, i.e. $$ \mathcal{L}_X(g) = 2 Ric_g $$ for $X=\nabla f$ for some function $f$. Then, let $\Phi_t$ denote the flow of $-X$. You may check that $h(t) :=\Phi^*_tg$ is a solution to the Ricci flow. Then, the Ricci curvature of $h(t)$ is $$ Ric_{h(t)} = Ric_{\Phi_t^*g} = \Phi_t^*Ric_g. $$ Thus, $$ \frac{d}{dt} Ric_{h(t)} = -\mathcal{L}_X Ric_g $$ Now, relating $\mathcal{L}_X Ric_g$ to $D_X Ric_g$ in the usual way: \begin{align*} \mathcal{L}_X Ric(V,W) & = X(Ric(V,W))-Ric([X,V],W)-Ric(V,[X,W])\\ & = D_X Ric(V,W) +Ric(D_XV-[X,V],W)+Ric(V,D_X W -[X,W])\\ & = D_X Ric(V,W) +Ric(D_VX,W)+Ric(V,D_WX) \end{align*} and using the formula for $\frac{d}{dt} Ric_{h(t)}$ under the Ricci flow, you may find an equation for $D_X Ric$ (for example, it is the equation found in the paper of Brendle's I linked above; I've matched his sign conventions with $X$).

EDIT: I'm going to work out the computation: I have specialized to gradient solitons (I'm not sure what happens if you drop this assumption).

From Topping's notes http://homepages.warwick.ac.uk/~maseq/topping_RF_mar06.pdf, we know that \begin{align*} \frac{d}{dt} Ric_{h(t)}(V,W) & = \Delta Ric_{h(t)} (V,W) - 2<Ric_{h(t)}(V),Ric_{h(t)}(W)>_{h(t)} \\ & \qquad + 2<Rm_{h(t)}(X,\cdot,V,\cdot),Ric_{h(t)}>_{h(t)}. \end{align*} If we specialize to $t=0$, then this gives us an equation for $\frac {d}{dt}Ric_{h(t)}|_{t=0}$ in terms of $g$-quantities.

On the other hand, notice that (here, we're using the gradient assumption) $$ DX = D^2f = \frac 12 \mathcal{L}_X g = Ric_g. $$ Thus $$ D_V X= Ric_g(V). $$

Now, you can put the $\frac{d}{dt}Ric_{h(t)}|_t=0$ computation together with the $\mathcal{L}_XRic_g$ computation to see \begin{align*} D_X Ric_g (V,W) & = \mathcal{L}_X Ric(V,W) -Ric_g(D_VX,W)-Ric_g(V,D_WX)\\ & = -\frac{d}{dt} Ric_{h(t)}|_{t=0}(V,W)-2<Ric_g(V),Ric_g(W)>\\ & = \Delta Ric_{g} (V,W) + 2<Rm_{g}(X,\cdot,V,\cdot),Ric_{g}>_{g}. \end{align*}

This is the equation that Brendle uses in the linked article.

By the way, you can simplify the above argument by using the Uhlenbeck trick (if you work through the details of what I did above, you'll see that there is a good deal of cancellation, which you can exploit by using $D_{\frac{d}{dt}}$ instead of the time derivative above. See, e.g. ch 6 of this book: http://download.springer.com/static/pdf/28/bok%253A978-3-642-16286-2.pdf?auth66=1385658655_8c41399e85b51a3addd04347ad203f0d&ext=.pdf among many other places.