the Levi-Civita connection on a product of Riemannian manifolds

The problem of verifying the Leibniz rule is closely related to your question as to what extent the formula given actually gives a well-defined map.

The point is that any vector field on $M_1\times M_2$ can locally be written as a linear combination of vector fields on $M_1$ and $M_2$, with the coeffients being functions on the product $M_1\times M_2$. The formula given should then be extended to all vector fields by assuming the Leibniz rule. (In short, the candidate connection on the product will satisfy the Leibniz rule by definition.)

You then have to check that what you have is well-defined, torsion-free, and compatible with the product metric. These should all be straightforward exercises, though.

I think that this can be solved by using the fact that the canonical maps from M1 and M2 into their product are immersions, and then applying exercise 3 of chapter 6.

Let $p\in M_1, q\in M_2.$ Observe that the canonical maps $M_1\rightarrow M_1\times \{q\}$ and $M_2\rightarrow \{p\}\times M_2$ are immersions of $M_1$ and $M_2$ into $M_1\times M_2.$ Next, notice that metrics induced on these immersed submanifolds are the same as their original metrics. Now, let $X_1,Y_1\in \chi (M_1\times\{q\})$ and $X_2,Y_2\in \chi(\{p\}\times M_2)$

Now, let $\overline\nabla$ be the Levi-Civita connection on $M_1\times M_2,$ by exercise 3 of chapter 6 we get,

$$\nabla^{1}_{X_1}Y_1=\big(\overline\nabla_{\overline{X}}\overline{Y}\big)^{Tan_{M_1}}$$ where $\overline{X}$ and $\overline{Y}$ are extensions of $X_1$ and $Y_1.$

We get a similar formula for $\nabla^{2}_{X_2}Y_2.$

Now let us define $\overline{X}$ and $\overline{Y}$ by;

$\overline{X}(a,b)=X_1(a,q)+X_2(p,b)$

$\overline{Y}(a,b)=Y_1(a,q)+Y_2(p,b)\quad \forall (a,b)\in M_1\times M_2.$

This gives us \begin{align*} \nabla^{1}_{X_1}Y_1+ \nabla^{2}_{X_2}Y_2&=\big(\overline\nabla_{\overline{X}}\overline{Y}\big)^{Tan_{M_1}}+\big(\overline\nabla_{\overline{X}}\overline{Y}\big)^{Tan_{M_2}} \\ &=\overline{\nabla}_{\overline{X}}\overline{Y}\end{align*} as required.

This equality is obviously not true in general. To see it, take $M=N=(0,+\infty)$. On the hand, $$\nabla_{(xe_1 + ye_2)}(-ye_1+xe_2)=-ye_1+xe_2,$$ on the other hand, $$\nabla_{xe_1}(-ye_1) = \nabla_{ye_2}(xe_2) = 0.$$ It happens because $-ye_1+xe_2$ is not a "decomposable" vector fields. Let's say that $X$ is decomposable if $X(p,q) = X_1(p) + X_2(q)$, $p \in M, q \in N,$ for $X_1 \in \mathfrak{X}(M), X_2 \in \mathfrak{X}(N).$

To be more precise, embed $M \hookrightarrow{} M \times \{q\} \hookrightarrow M \times N$ for every $q \in N$. The collection of pushforwards of $X$, still denoted by $X$, is a smooth vector field on $M\times N$.

What is true is that, if X and Y are decomposable, then $$\nabla_{Y}X = \nabla^1_{Y_1}X_1 + \nabla^2_{Y_2}X_2,$$ which is equivalent to prove that $$\nabla^1_{Y_1}X_2=\nabla^1_{Y_2}X_1=\nabla^1_{Y_2}X_2 = 0.$$ But $X_1$ does not vary on $N$ direction, so $\nabla_{Y_2}X_1 = 0. $The same for $\nabla_{Y_1}X_2$. It remains to show the last equality. Take local coordinates $(x_1, \dots, x_m)$ on $M^m$ and $(y_1, \dots, y_n)$ on $N^n$. Using Koszul formula, we have

$$2\langle \nabla_{Y_2}X_2, \frac{\partial}{\partial x^i}\rangle = X_2 \underbrace{\langle \frac{\partial}{\partial x^i},Y_2\rangle}_{=0} +Y_2 \underbrace{\langle X_2,\frac{\partial}{\partial x^i} \rangle}_{=0} -\frac{\partial}{\partial x^i}\underbrace{\langle X_2,Y_2\rangle}_{\rm{don't \ vary \ on \ M}} \\ -\langle[X_2,\frac{\partial}{\partial x^i}],Y_2 \rangle -\langle[Y_2,\frac{\partial}{\partial x^i}],X_2 \rangle- \langle\underbrace{[X_2,Y_2]}_{\in \mathfrak{X}(N)},\frac{\partial}{\partial x^i} \rangle$$ Writting $X_2(x,y) = X_2(y) = \phi^j(y)\frac{\partial}{\partial y^j}$, $$ [X_2,\frac{\partial}{\partial x^i}] = \phi^j[\frac{\partial}{\partial y^j}, \frac{\partial}{\partial x^i}] - \frac{\partial \phi^j}{\partial x^i}\frac{\partial}{\partial y^j} =0, $$ which proves the desired formula.

Observe that this is enough to prove the remaining itens of the exercise.