= Solution
A <Levi-Civita connection> on a <Riemannian manifold> is a <covariant derivative> $D$ that is a <torsion-free connection>, $D_XY-D_YX=[X,Y]$, and a <metric connection>, $Xg(Y,Z)=g(D_XY,Z)+g(Y,D_XZ)$. A <covariant derivative> is $C^\infty(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+fD_XY$.
Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the <Koszul formula>
$$
2g(D_XY,Z)=Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)-g(X,[Y,Z])+g(Y,[Z,X])+g(Z,[X,Y]).
$$
Since $g$ is nondegenerate, this determines $D_XY$ uniquely. For existence, use the right-hand side to define $D_XY$: expansion of the <Lie bracket of vector fields> shows it is $C^\infty(M)$-linear in $Z$, hence defines a smooth one-form, which the <musical isomorphism> converts to a smooth <vector field>. The same expansion shows $D_{fX}Y=fD_XY$ and $D_X(fY)=X(f)Y+fD_XY$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a <Levi-Civita connection>. Equivalently its coefficients in a <manifold chart> are the <Christoffel symbols>
$$
\Gamma^k_{ij}=\frac12 g^{k\ell}(\partial_i g_{j\ell}+\partial_j g_{i\ell}-\partial_\ell g_{ij}).
$$
The intrinsically defined <Koszul formula> ensures these local expressions fit together. This proves the <fundamental theorem of Riemannian geometry>.
The <product Riemannian metric> on $M\times N$ is
$$
g_{(m,n)}((u,v),(u',v'))=(g_M)_m(u,u')+(g_N)_n(v,v').
$$
The two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_XY$ against lifted local frame fields from both factors, which together span each product tangent space.
If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the <Koszul formula> is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_XY$ is orthogonal to both spanning frame families, and positive definiteness yields
$$
\boxed{D_XY=0.}
$$
This calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.
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