I'll use the same notations as yours.
In Riemann normal coordinates, the point is $x^{\alpha}=\lambda\, n^{\alpha}$, the tangent vector is $u^{\alpha} = n^{\alpha}$, the deviation vectors are $N_{(\mu)}^{\alpha} = \lambda \,\delta_{(\mu)}^{\alpha}$.
The geodesic deviation equation is
\begin{equation}\tag{1}
\frac{\nabla^2 N_{(\mu)}^{\alpha}}{d\lambda^2} = {R^{\alpha}}_{\beta\gamma\delta} u^{\beta} u^{\gamma} N_{(\mu)}^{\delta}.
\end{equation}
Expanding Eq. (1) using the definition of covariant derivative
\begin{equation}\tag{2}
\frac{\nabla N_{(\mu)}^{\alpha}}{d\lambda} = \frac{dN_{(\mu)}^{\alpha}}{d\lambda} + \Gamma^{\alpha}_{\beta\gamma} u^{\beta} N_{(\mu)}^{\gamma},
\end{equation}
and the geodesic equation $d u^{\alpha}/d\lambda + \Gamma^{\alpha}_{\beta\gamma} u^{\beta} u^{\gamma} = 0$, we obtain
\begin{equation}\tag{3}
\frac{d^2 N_{(\mu)}^{\alpha}}{d\lambda^2} + 2 \Gamma^{\alpha}_{\gamma\delta} u^\gamma \frac{dN_{(\mu)}^{\delta}}{d\lambda} + \Bigl( \Gamma^{\alpha}_{\gamma\delta,\beta} - \Gamma^{\alpha}_{\mu\delta} \Gamma^{\mu}_{\beta\gamma} + \Gamma^{\alpha}_{\beta\mu} \Gamma^{\mu}_{\gamma\delta} - {R^\alpha}_{\beta\gamma\delta} \Bigr) u^{\beta} u^{\gamma} N_{(\mu)}^{\delta} = 0.
\end{equation}
Substituting $n^{\alpha}$ and $N_{(\mu)}^{\alpha}$ into the above equation, we find
\begin{equation}\tag{4}
2 \Gamma^{\alpha}_{\gamma\delta} n^{\gamma} + \lambda \Bigl( \Gamma^{\alpha}_{\gamma\delta,\beta} - \Gamma^{\alpha}_{\mu \delta} \Gamma^{\mu}_{\beta\gamma} + \Gamma^{\alpha}_{\beta\mu} \Gamma^{\mu}_{\gamma\delta} - {R^{\alpha}}_{\beta\gamma\delta} \Bigr) n^{\beta} n^{\gamma} = 0.
\end{equation}
We expand $\Gamma^{\alpha}_{\gamma\delta}$ in powers of $\lambda$:
\begin{equation*}
\Gamma^{\alpha}_{\gamma\delta} = \Gamma^{\alpha}_{\gamma\delta} \big|_{\mathcal{P}_0} + \lambda \Gamma^{\alpha}_{\gamma\delta,\mu} u^{\mu} + O(\lambda^2) = \lambda \Gamma^{\alpha}_{\gamma\delta,\beta} n^{\beta} + O(\lambda^2).
\end{equation*}
Substituting this expression back into (4), dividing through by $\lambda$ and then evaluating on $\mathcal{P}_0$, we arrive at
\begin{equation*}
\bigl( 3 \Gamma^{\alpha}_{\gamma\delta,\beta} - {R^{\alpha}}_{\beta\gamma\delta} \bigr)\big|_{\mathcal{P}_0} n^{\beta} n^{\gamma} = 0.
\end{equation*}
After symmetrizing in the indices $\beta$ and $\gamma$, we get
\begin{equation}\tag{5}
\bigl( \Gamma^{\alpha}_{\gamma\delta,\beta} + \Gamma^{\alpha}_{\beta\delta,\gamma} \bigr) \big|_{\mathcal{P}_0} = \frac{1}{3} \bigl( {R^{\alpha}}_{\beta\gamma\delta} + {R^{\alpha}}_{\gamma\beta\delta} \bigr) \big|_{\mathcal{P}_0}.
\end{equation}
By cycling the indices $(\gamma\delta\beta)$ to get two other equations $A$ and $B$, adding $A$ to and substracting $B$ from (5), we finally get
\begin{equation}\tag{6}
\Gamma^{\alpha}_{\beta\delta,\gamma} \big|_{\mathcal{P}_0} = \frac{1}{3} \bigl( {R^{\alpha}}_{\beta\gamma\delta} + {R^{\alpha}}_{\delta\gamma\beta} \bigr) \big|_{\mathcal{P}_0}.
\end{equation}
If you like, you may also write it as
\begin{equation}\tag{7}
\Gamma^{\alpha}_{\beta\gamma,\delta} \big|_{\mathcal{P}_0} = - \frac{1}{3} \bigl( {R^{\alpha}}_{\beta\gamma\delta} + {R^{\alpha}}_{\gamma\beta\delta} \bigr) \big|_{\mathcal{P}_0}.
\end{equation}
