It seems to me that people usually include the $\mathbb{R}$-linearity as part of the definition (correct me if I'm wrong); i.e. define $\nabla$ on the simple tensors by the product rule above, and then extend it linearly over $\mathbb{R}$ to the whole space $C^{\infty}(E\otimes F)$.
Actually, due to the fact that $C^{\infty}(E\otimes F)\simeq C^{\infty}(E)\otimes_{C^{\infty}(M)}C^{\infty}(F)$ which OP has mentioned in the comment above, it follows that additivity (i.e.
\begin{align}
\nabla_X(s\otimes t+s'\otimes t')=\nabla_X(s\otimes t)+\nabla_X(s'\otimes t')
\end{align}
which is weaker than $\mathbb{R}$-linearity) is sufficient.
With this in mind, here is an alternative proof, which is essentially a direct computation, though perhaps it may look less elegant.
Let $A\in C^{\infty}(E\otimes F)$. Such $A$ can be expressed as a finite sum of simple tensors, though not uniquely in general. So we want to show that if it can be written in the following two ways
\begin{align}
A=\sum_is_i\otimes t_i=\sum_j\tilde{s}_j\otimes\tilde{t}_j
\end{align}
(the number of summands in these two expressions may be different in general), then
\begin{align}
\nabla_X\left(\sum_is_i\otimes t_i\right)
=\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right) & & (*)
\end{align}
Let $(e_{\alpha})$ and $(\epsilon_{\beta})$ be smooth local frames for $E$ and $F$ respectively. Then one can write
\begin{align}
s_i=s_i^{\alpha}e_{\alpha},\qquad
t_j=t_j^{\beta}\epsilon_{\beta},\qquad
\tilde{s}_i=\tilde{s}_i^{\alpha}e_{\alpha},\qquad
\tilde{t}_j=\tilde{t}_j^{\beta}\epsilon_{\beta}
\end{align}
(Einstein summation convention is assumed.) Then we will have
\begin{align}
\sum_is_i\otimes t_i
=\left(\sum_is_i^{\alpha}t_i^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta},
\qquad
\sum_j\tilde{s}_j\otimes\tilde{t}_j
=\left(\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
Since $\{e_{\alpha}\otimes\epsilon_{\beta}\}$ is a smooth local frame for $E\otimes F$, by uniqueness of local components we must have
\begin{align}
\sum_is_i^{\alpha}t_i^{\beta}=\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta} & & (1)
\end{align}
Taking exterior derivative we then have
\begin{align}
\sum_i\left(t_i^{\beta}ds_i^{\alpha}+s_i^{\alpha}dt_i^{\beta}\right)
=\sum_j\left(\tilde{t}_j^{\beta}d\tilde{s}_j^{\alpha}
+\tilde{s}_j^{\alpha}d\tilde{t}_j^{\beta}\right) & & (2)
\end{align}
Now let $\omega_{\alpha}^{\beta}$ and $\theta_{\alpha}^{\beta}$ be the connection 1-forms of $\nabla^E$ and $\nabla^F$ respectively, so that e.g.
\begin{align}
\nabla_X s_i=\left[Xs_i^{\alpha}+s_i^{\beta}\omega_{\beta}^{\alpha} (X)\right]e_{\alpha} & & (3)
\end{align}
and similar identities hold for $\nabla_Xt_i$, $\nabla_X\tilde{s}_j$ and $\nabla_X\tilde{t}_j$. Then we can compute
\begin{align}
&\nabla_X\left(\sum_is_i\otimes t_i\right) \\
%%%
&=\sum_i\left(\nabla_Xs_i\otimes t_i+s_i\otimes\nabla_Xt_i\right) \\
%%%
&=\underbrace{\left(\sum_i\left((Xs_i^{\alpha})t_i^{\beta}
+s_i^{\alpha}(Xt_i^{\beta})\right)\right)}_{=:I}e_{\alpha}\otimes\epsilon_{\beta}
+\underbrace{\left(\sum_is_i^{\gamma}t_i^{\beta}\right)}_{=:II}
\omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta}
+\underbrace{\left(\sum_is_i^{\alpha}t_i^{\gamma}\right)}_{=:III}
\theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
where the first step follows by additivity and product rule, while the second step is obtained by substitution of (3) and rearranging the terms.
Now $\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)$ will have the same expression, except that all of the $s_i,t_i$ are replaced by $\tilde{s}_j,\tilde{t}_j$; says
\begin{align}
\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)
=\tilde{I}\cdot e_{\alpha}\otimes\epsilon_{\beta}
+\tilde{II}\cdot \omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta}
+\tilde{III}\cdot \theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
Then:
- By (2), we have $I=\tilde{I}$.
- By (1), we have $II=\tilde{II}$ and $III=\tilde{III}$.
Hence, (*) holds as desired.
A possible way to prove this is to remember that the LC connection is the unique torsion free connexion for which the metric tensor is parallel.
The fact that your formula gives a connexion is obvious.
To check that it is torsion free, note that $g_N(\tilde \nabla _X Y, Z)= g(\nabla _X Y, Z)$ for every triple of tangent vector fields on $N$
To check that it preserves the induced tensor metric let $X,Y,Z$ three tangent vector fields on $N$. We can extend these fields on $M$ to compute :
$(\tilde \nabla _X g) (X,Y)= X. g(Y,Z)-g(\tilde \nabla _X Y, Z)- g(Y, \tilde \nabla _XZ)=X. g(Y,Z)-g( \nabla _X Y, Z)- g(Y, \nabla _XZ) $
Best Answer
For 1, it is an implicit definition. It says that given $\nabla$ (acting on vector fields on $M$ and sections of $E$) there exists a unique operator $\tilde{\nabla}:TM\times\Gamma(E)\to E$ (which they have still denoted as $\nabla$) such that for each $p\in M$,
So, the definition of $\tilde{\nabla}$ is as a unique map satisfying such and such property. The point is that you have to ensure well-definition of $\tilde{\nabla}$ because there are infintely many vector fields $X$ whose value at a particular point $p$ is equal to a given tangent vector $h\in T_pM$. The reason for the existence and uniqueness of such a $\tilde{\nabla}$ is simply because $\nabla$ is $C^{\infty}(M)$-linear in the vector field slot where $X$ goes.
To check that these depend on the section $s$ only through its values along a given curve with tangent $h$, simply write down the local coordinate formula involving the Christoffel symbols; I think that’s the fastest way to go (see e.g the second half of my answer here… I only talk about connections in the tangent bundle, but the same idea goes through for every vector bundle).