Let $E \to M$ be a vector bundle (with fibres $V$) over a smooth manifold $M$. Define the covariant derivative $\nabla$ as a map $$\nabla : C^\infty(M,E) \to C^\infty(M,E \otimes T^*M),$$ where $C^\infty(M,E)$ denotes the smooth sections of $E$. This map is linear and satisfies a product rule, mainly $$\nabla(f\sigma) = f\nabla \sigma + \sigma \otimes df, \qquad f \in C^\infty(M)\text{ and } \sigma \in C^\infty(M,E).$$ I want to show that a covariant derivative always exists locally. For that, consider an open set $U \subset M$ such that there is a bundle isomorphism $E|_U \simeq U \times V$. Thus any section on $U$ can be written as a map $\sigma : x \mapsto (x, s(x))$, where $s : U \to V$ is smooth. Then, define $\nabla \sigma$ as the map $x \mapsto (x, s_*|_x)$.
First of all, I need to check that $\nabla \sigma \in C^\infty(M, (U \times V) \otimes T^*M)$. This amounts to showing that, for all $x \in U$,
$$s_*|_x \in (U \times V)|_x \otimes T_x^*M \simeq V \otimes T_x^*M.$$
First question: I know about the canonical isomorphism $L^* \otimes M \simeq \mathcal L(L, M)$, so what's the motivation behind defining the covariant derivative to take values in sections of $E \otimes T^*M$ and not sections of $T^*M \otimes E$?
With that aside, one can see that
$$s_*|_x : T_xM \to V \Rightarrow s_*|_x \in \mathcal L(T_xM, V) \simeq T_x^*M \otimes V \simeq V \otimes T_x^*M,$$
where the last isomorphism is not canonical.
Now, we need only check the product rule. If $f \in C^\infty(U)$, I'd want something like $$(fs)_*|_x = f(x)s_*|_x + df|_xs(x).$$ The first term on the right hand side makes sense, and is clearly in $V \otimes T_x^*M$. Also, $df|_x$ is a linear functional on $T_xM$, so it's clearly in $T_x^*M$. Obviously, $s(x) \in V \simeq (U \times V)|_x$. I can definitely see some sort of product, and I conjecture that $df|_x \otimes s(x)$ is the map that takes $v \in T_xM$ to $df|_x(v)s(x)$.
Second question: This is not quite right. If $U = \mathbf{R}^n$ and $V = \mathbf{R}^m$, then $D(fs)(x) = Df(x)s(x) + f(x)Ds(x)$. The first term of the right hand side is only defined for $n = m$. That would imply that the fiber dimension of $E$ would have to be equal to the dimension of $M$. So, what exactly is the product I want between $df|_x$ and $s(x)$ and how exactly would it arrive naturally? I guess what I'm having trouble with is that I don't know what elements of $E \otimes T^*M$ look like.
EDIT: The product rule I used above is incorrect. In fact, one has $$D(fs)(x) = f(x)Ds(x) + P(Df(x), s(x)),$$ where $P(Df(x),s(x))$ is the $m\times n$ matrix whose i-th row is $Df(x)s_i(x)$, and I can see that this is $Df(x) \otimes s(x)$.
The product rule I used above is incorrect. In fact, one has $$D(fs)(x) = f(x)Ds(x) + P(Df(x), s(x)),$$ where $P(Df(x),s(x))$ is the $m\times n$ matrix whose i-th row is $Df(x)s_i(x)$, and I can see that this is $Df(x) \otimes s(x)$.