I have the manifold $ M \times N $, where $ M, N $ are 2-dim manifolds, with coordinates $ (x_1, x_2) $ and $ (x_3, x_4) $ respecrively. On $ M\times N $ i have the metric
$ g(\mathbb x, \mathbb x') = \begin{bmatrix} \phi(\mathbb x) & 0 \\ 0 & \psi(\mathbb x') \end{bmatrix} $
So $ ds^2_{M\times N} = ds^2_M+ds^2_N $
$ ds^2_{M\times N}(x_1, x_2, x_3, x_4) = $
$ = \phi_{11} dx_1^2 + \phi_{22}dx_2^2+2\phi_{12}dx_1dx_2 + $
$ + \psi_{33}dx_3^2+\psi_{44}dx_4^2 + 2\psi_{34}dx_3dx_4 $
The curvature is $ R_{ijkl}\\ $
I know by symmetry that $ R_{11ks} = R_{22ks} = R_{33ks} = R{44ks} = 0 $
$ R_{ij11} = R_{ij22}=R_{ij33} = R_{ij44} = 0 $
Then
$ R_{ijks} = -R_{jiks} $
$ R_{ijks} = -R_{ijsk} $
So $ R_{jiks} = R_{ijsk} $
which implies
$ R_{ijks} = R_{ksij} = R_{ksji} $
Now, if $ i=k $ and $ j=s $,
$ R_{ijij} = -R_{ijji} = R_{jiji} = - R_{jiij} $
So the first 6 independent symbols to calculate are
$ R_{1212} \quad R_{1313} \quad R_{1414} $
$ R_{2323} \quad R_{2424} \quad R_{3434} $
The other 14 independent symbols are $ R_{1213} \quad R_{1214} \quad R_{1223}$
$ R_{1224} \quad R_{1234} \quad R_{1314} $
$ R_{1323} \quad R_{1324} \quad R_{1334} $
$ R_{1423} \quad R_{1434} \quad R_{2324} $
$ R_{3423} \quad R_{3424} $
I am not sure whether all those symbols i wrote are the correct independent ones. I know the formula for Christoffel symbols, then I need to calculate all those 20 symbols, but I know most of them will be zero. Am I doing good? Because i think i am overdoing and making mistakes.
EDIT ($\textbf{about Christoffel's symbols and sectional curvature}$):
Considering the symmetries $ \Gamma_{aij} = \Gamma_{aji} $, we need to calculate the following Christoffel symbols:
$ \Gamma_{111} \quad \Gamma_{112} \quad \Gamma_{122} \quad \Gamma_{211} \quad \Gamma_{212} \quad \Gamma_{222} $
$ \Gamma_{333} \quad \Gamma_{334} \quad \Gamma_{344} \quad \Gamma_{433} \quad \Gamma_{434} \quad \Gamma_{444} $
$ \Gamma_{111} = \partial_1g_{11}/2 \quad \Gamma_{112} =\partial_2g_{11}/2\quad \Gamma_{122} = (2\partial_2g_{12} - \partial_{1}g_{22})/2 $
$ \Gamma_{211} = (2\partial_1g_{21} - \partial_2g_{11})/2 \quad \Gamma_{212} = \partial_1g_{22}/2 \quad \Gamma_{222} = \partial_2 g_{22} /2 $
$ \Gamma_{333} = \partial_3g_{33}/2 \quad \Gamma_{334} = \partial_4g_{33}/2 \quad \Gamma_{344} = (2\partial_4g_{34} - \partial_{3}g_{44})/2 $
$ \Gamma_{433} = (2\partial_3g_{43} - \partial_4g_{33})/2 \quad \Gamma_{434} = \partial_3g_{44}/2 \quad \Gamma_{444} = \partial_4 g_{44}/2 $
And what about the sectional curvature $S$? Being $M, N$ bidimensional, we only have one tangent plane, so one sectional curvature. In dimension 2, we know that the Ricci tensor coincides with the sectional curvature.
EDIT II ($\textbf{forgot to add metric symbols}$):
For $M$:
$ S(e_1, e_2) = \frac{Ric(e_1, e_1) + Ric(e_2, e_2)}{2} $
$ Ric(e_1, e_1) = g(R(e_1, e_1)e_1,e_1) + g(R(e_1, e_2)e_2, e_1) = $
$ = R^1_{111}g_{11} + R^2_{111}g_{21} + R^1_{122}g_{11} + R_{122}^2g_{21} $
$ Ric(e_2, e_2) = $
$ = g(R(e_2, e_1)e_1,e_2) + g(R(e_2, e_2)e_2, e_2) $
$ = R^1_{211}g_{12} + R^2_{211}g_{22} + R^1_{222}g_{12} + R^2_{222}g_{22} = $
This implies $ S(M) = $
$ = \frac{g_{11}}{2}(R^1_{111} + R^1_{122}) + \frac{g_{22}}{2}(R^2_{211} + R^2_{222}) $
$ +\frac{g_{12}}{2}(R^2_{111} + R^2_{122} + R^1_{222} + R^1_{211}) $
Similarly for $ N $.
Let $R^{M}$ be the Riemann curvature of $M$, and $R^{N}$ be the Riemann curvature of $N$. Since you have just a product manifold where the $(x_1,x_2)$ and $(x_3,x_4)$ coordinates are independent, the resulting Riemann curvature has the property:
$$R_{abcd} (x,x')=\\= \begin{cases} R^{M}_{abcd}(x)\quad \text{if} \quad \{a,b,c,d\}\subset\{1,2\} \\ R^{N}_{abcd}(x')\quad \text{if} \quad \{a,b,c,d\}\subset\{3,4\} \\ 0 \quad \text{in any other case (that is,} \\ \quad\; \text{ when indices come from both subspaces.)} \end{cases}$$
You still need to compute the Riemann curvature of the $2$-dimensional $M$ and $N$ submanifolds, but it's easier than working in the full $4$-dimensional manifold.
An informal justification for the above would be that parallel transport of a vector from the $M$ component of the tangent space along a curve formed in the $N$ component leaves it unchanged. Any change in the $M$-components of a vector must come from movement along the $M$-subspace.