Derivation of Gauss-Codazzi type equation (Ricci relation) I am following Padmanabhan's book Gravitation for the particular derivation. The derivation goes as follows,
\begin{align}
    R_{abst}n^t&=\nabla_a\nabla_b n_s-\nabla_b\nabla_a n_s=\nabla_a(-K_{bs}-n_b a_s)-\nabla_b(-K_{as}-n_a a_s)\nonumber\\
    &=-\nabla_a K_{bs}-n_b\nabla_a a_s-a_s\nabla_a n_b+\nabla_b K_{as}+n_a \nabla_b a_s+a_s\nabla_b n_a
\end{align}
Contracting again with a normal vector,
\begin{align}
    R_{abst}n^b n^t &=-n^b \nabla_a K_{bs}+\nabla_a a_s+n^b \nabla_b K_{as}+n_a n^b \nabla_b a_s+a_s n^b\nabla_b n_a\nonumber\\
    &=-n^b \nabla_a K_{bs}+(\delta^r_a+n_an^r)\nabla_r a_s+n^b \nabla_b K_{as}+a_s n^b\nabla_b n_a\nonumber\\
    &=-n^b \nabla_a K_{bs}+h^r_a \nabla_r a_s+n^b \nabla_b K_{as}+a_s a_a
\end{align}
Now we project the free indices onto $\Sigma(t)$,
\begin{align}
    R_{abst}h^a_m n^b h^s_n n^t
    &=-n^b h^a_m h^s_n \nabla_a K_{bs}+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n
\end{align}
Here we have used the fact that $n_a a^a=0\implies h^a_b a_a=a_b$. Moreover, from $K_{ab}n^b=0$ we have, $n^b\nabla_m K_{ab}=-K_{ab}\nabla_m n^b$, thus we have,
\begin{align}
    R_{abst}h^a_m n^b h^s_n n^t
    &=K_{bn} h^a_m \nabla_a n^b+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber\\
    &=-K_{bn} K^b_m+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber
\end{align}
Now,
\begin{align}
    \mathcal{L}_{\mathbf{n}}K_{as}=n^b \nabla_b K_{as}+K_{al}\nabla_s n^l+K_{ls}\nabla_a n^l.
\end{align}
\begin{align}\label{eq:curavaturetwonormal}
    R_{abst}h^a_m n^b h^s_n n^t
    &=K_{bn} h^a_m \nabla_a n^b+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber\\
    &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as}-K_{al}\nabla_s n^l-K_{ls}\nabla_a n^l)+a_m a_n\nonumber\\
    &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})-K_{ml} h^s_n\nabla_s n^l-K_{ln}h^a_m \nabla_a n^l+a_m a_n\nonumber\\
    &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})+K_{ml} K^l_n+K_{nl}K^l_m+a_m a_n\nonumber\\
    &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})+2K_{bn}K^b_m+a_m a_n\nonumber\\
&=K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (\mathcal{L}_{\mathbf{n}}K_{as})+a_m a_n
\end{align}
The extrinsic curvature is defined as $-K_{mn}=h^a_m\nabla_an_n$.
\begin{align}
    \mathcal{L}_{\mathbf{n}}K_{mn}&=\mathcal{L}_{\mathbf{n}}(h^a_m h^a_n K_{as})\nonumber\\
    &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})+h^a_m K_{as} (\mathcal{L}_{\mathbf{n}} h^s_n)+h^s_n K_{as} (\mathcal{L}_{\mathbf{n}}h^a_m)\nonumber\\
    &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})-2h^a_m K_{as} K^s_n-2h^s_n K_{as} K^a_m\nonumber\\
    &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})-4 K_{nb} K^b_m
\end{align}
Thus, we have,
\begin{align}
    h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})=\mathcal{L}_{\mathbf{n}}K_{mn}+4 K_{nb} K^b_m
\end{align}
Putting this into the original equation, we obtain,
\begin{align}
    R_{abst}h^a_m n^b h^s_n n^t
    &=5K_{bn} K^b_m+D_m a_n+\mathcal{L}_{\mathbf{n}}K_{mn}+a_m a_n
\end{align}
Which doesn't seem right.
In the book, however, we have something like this

Could someone please explain what I am doing wrong and how to obtain the correct results?
 A: Try this following sequence of derivation instead.
\begin{align}
    R_{abst}n^t&=\nabla_a\nabla_b n_s-\nabla_b\nabla_a n_s=\nabla_a(-K_{bs}-n_b a_s)-\nabla_b(-K_{as}-n_a a_s)\nonumber\\
    &=-\nabla_a K_{bs}-n_b\nabla_a a_s-a_s\nabla_a n_b+\nabla_b K_{as}+n_a \nabla_b a_s+a_s\nabla_b n_a
\end{align}
Contracting with a normal vector,
\begin{align}
    R_{abst}n^b n^t &=-n^b \nabla_a K_{bs}+\nabla_a a_s+n^b \nabla_b K_{as}+n_a n^b \nabla_b a_s+a_s n^b\nabla_b n_a\nonumber\\
    &=-n^b \nabla_a K_{bs}+(\delta^r_a+n_an^r)\nabla_r a_s+n^b \nabla_b K_{as}+a_s n^b\nabla_b n_a\nonumber\\
    &=-n^b \nabla_a K_{bs}+h^r_a \nabla_r a_s+n^b \nabla_b K_{as}+a_s a_a
\end{align}
Projecting onto the hypersurface,
\begin{align}
    R_{abst}h^a_m n^b h^s_n n^t
    &=-n^b h^a_m h^s_n \nabla_a K_{bs}+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n
\end{align}
Using the fact, $n_a a^a=0\implies h^a_b a_a=a_b$. Moreover, $K_{ab}n^b=0\implies n^b\nabla_m K_{ab}=-K_{ab}\nabla_m n^b$, gives,
\begin{align}
    R_{abst}h^a_m n^b h^s_n n^t
    &=K_{bn} h^a_m \nabla_a n^b+D_m a_n+n^b h^a_m h^s_n \nabla_b K_{as}+a_m a_n\nonumber\\
    &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (n^b\nabla_b K_{as})+a_m a_n
\end{align}
Now,
\begin{align}
  h^a_m h^s_n\mathcal{L}_{\mathbf{n}}K_{as}&=h^a_m h^s_n\left(n^b\nabla_b K_{as}+K_{bs}\nabla_a n^b+K_{ab}\nabla_s n^b\right)\nonumber\\
  &=h^a_m h^s_n\left(n^b\nabla_b K_{as}\right)-K_{bn}K_m^b-K_{mb}K^b_n
\end{align}
Also,
\begin{align}
  \mathcal{L}_{\mathbf{n}}h^a_b&=\mathcal{L}_{\mathbf{n}}\left(\delta^a_b+n^a n_b\right)\nonumber\\
  &=\mathcal{L}_{\mathbf{n}}\delta^a_b+\mathcal{L}_{\mathbf{n}}(n^a n_b)\nonumber\\
  &=n^m\nabla_m \delta^a_b-\delta^m_b\nabla_m n^a+\delta^a_m\nabla_b n^m+n^m\nabla_m(n^a n_b)-n^m n_b\nabla_m n^a+n^a n_m\nabla_b n^m\nonumber\\
  &=n^m n^a\nabla_m n_b\nonumber\\
  &=n^a a_b
\end{align}
Then,
\begin{align}
   \mathcal{L}_{\mathbf{n}}K_{mn}&=\mathcal{L}_{\mathbf{n}}(h^a_m h^s_n K_{as})\nonumber\\
    &=h^a_m h^s_n (\mathcal{L}_{\mathbf{n}} K_{as})+h^a_m K_{as} (\mathcal{L}_{\mathbf{n}} h^s_n)+h^s_n K_{as} (\mathcal{L}_{\mathbf{n}}h^a_m)\nonumber\\
    &=h^a_m h^s_n\left(n^b\nabla_b K_{as}\right)-K_{bn}K_m^b-K_{mb}K^b_n-h^a_m K_{as} n^s a_n-h^s_n K_{as} n^a a_m\nonumber\\
    &=h^a_m h^s_n\left(n^b\nabla_b K_{as}\right)-K_{bn}K_m^b-K_{mb}K^b_n\nonumber
\end{align}
Putting all these together, we have,
\begin{align}
   R_{abst}h^a_m n^b h^s_n n^t
    &=-K_{bn} K^b_m+D_m a_n+ h^a_m h^s_n (n^b\nabla_b K_{as})+a_m a_n\nonumber\\
    &=-K_{bn} K^b_m+D_m a_n+ \mathcal{L}_{\mathbf{n}}K_{mn}+2K_{bn}K^b_m+a_m a_n\nonumber\\
    &=K_{bn} K^b_m+D_m a_n+ \mathcal{L}_{\mathbf{n}}K_{mn}+a_m a_n\nonumber
\end{align}
This is the desired expression.
