Why is connection a map from $\Gamma(E)$ to $\Omega^1\otimes\Gamma(E)$? On the site Vector Bundle Connection, it gives two definitions of a connection.
One is view a connection as a linear map from a section of $E\otimes TM$ to a section of $E$:
$$
D:\Gamma(E\otimes TM)\rightarrow\Gamma(E)
$$
I can understand this definition, thinking a connection as a directional directive:
$$
v\otimes w\mapsto D_vw
$$
However, I cannot understand the other definition, and which is seemingly more common:
$$
D:\Gamma(E)\mapsto\Gamma(E\otimes T^*M)=\Gamma(E)\otimes\Omega^1
$$
Can anyone explain to me how such map works? Given a vector in $\Gamma(E)$, what is the image of it?
In addition, in the site above there is an example about the connection in a trivial bundle, saying that 

Any connection on the trivial bundle $E=M\times\mathbb{R}^k$is of the form $\nabla s=ds+s\otimes\alpha$ where $\alpha$ is a one-form with value in Hom($E,E$).

However, I don't understand. I think $ds$ is an element in the dual bundle of $E$, but $s\otimes\alpha$ is not, although I cannot even point out to what space $s\otimes\alpha$ belongs, how can they be added?
 A: This definition tells you that a connection takes a vector field in $\Gamma(E)$ and produces an $E$-valued one-form. An $E$-valued one form (a section of $\Gamma(E)\otimes \Omega^1$) is a beast that takes in a vector field and produces a section of $E$ in a tensorial fashion (i.e., it's linear in smooth functions).
So in particular, if you have a covariant derivative $v\otimes s\to D_vs$, you should think of this definition as a "differential":
$$ s \mapsto Ds: \big( v \mapsto D_vs\big).$$
As to the local coordinate expression, $\alpha$ is a one-form with values in $\operatorname{Hom}(E,E)$. That is, it is an endomorphism-valued one-form: It takes in a vector at a point $p$ and produces a linear map $E_p\to E_p$. Think of $s\otimes\alpha$ as a multiplication operator containing Christoffel symbols in the same way that a covariant derivative can be locally expressed as $Dw = dw + \sum w_k\Gamma^k_{ij}$.
A: It's not really correct to think of connections as maps $\Gamma\left(E\otimes TM\right)\to\Gamma\left(E\right)$,
because the interplay between the $E$ and $TM$ inputs is not $C^{\infty}\left(M\right)$-multilinear.
More precisely, $s\otimes\left(fu\right)=\left(fs\right)\otimes u$,
so if the connection factored via the tensor product we should have
$D_{fu}s=D_{u}fs$; but $D_{fu}s=fD_{u}s$ is not equal to $D_{u}fs=\left(uf\right)s+f$$D_{u}s$
in general. Thus if you want to "keep the inputs together", you can't go any further than thinking of the connection as a map $\Gamma\left(E\right)\times\Gamma\left(TM\right)\to\Gamma\left(E\right)$.
Now, the other interpretation is indeed correct and common - given
a section $s\in\Gamma\left(E\right)$, the derivative of $s$ is a
$E$-valued one-form. For a section $s\in\Gamma\left(E\right)$, its
image is the map $Ds:\Gamma\left(TM\right)\to\Gamma\left(E\right):u\mapsto D_{u}s$.
This map is $C^{\infty}\left(M\right)$-linear, so we can in
fact interpret it as an element of $\Gamma\left(E\otimes T^{*}M\right)$.
