Reference for this question is the book Geometry of Differential forms by Morita, page no 191.
In last but one paragraph it says the following,
Suppose we are given an arbitrary bundle $\pi:E\rightarrow M$ with a connection $\nabla$. We think of $\nabla s$ as a $1$ form on $M$ whose value on vector field $X$ on $M$ is the section $\nabla_Xs$.
Connection on a vector bundle $\pi:E\rightarrow M$ is a bilinear map $\nabla:X(M)\times \Gamma(E)\rightarrow\Gamma(E)$ satisfying certain conditions where $X(M)$ is the collection of vector fields on $M$ and $\Gamma(E)$ is collection of sections of vector bundle.
I do not really understand the one that I have quoted above.
$1$ form on $M$ is a $\omega:M\rightarrow \Lambda(TM)$. They are saying $\nabla_s$ is a $1$ form on $M$. We need to say what is $\nabla_s:M\rightarrow \Lambda(TM)$.
Let $m\in M$ then, $\nabla_s(m):T_mM\rightarrow \mathbb{R}$. Let us fix some vector field $X$ on $M$. We need to define $\nabla_s(m)(X(m))$.
Suppose we are in trivial vector bundle, where sections are just smooth maps on $M$ we can define $\nabla_s(m)(X(m))$ as $\nabla_X(s)(m)\in \mathbb{R}$. I do not see how we can think of $\nabla_s$ as a $1$ form for a bundle which is not trivial.
I see that they are trying to relate connection on a vector bundle to be something that takes a section on $E$ and gives a section on some bundle, definitely not a section on $M$ as they have written. I fail to understand this.
A connection on a vector bundle is a map that takes a section on $M$ and gives a section on ..????
No. Morita is saying $\nabla s$ is an $E$-valued 1-form on $M$, i.e. a map $M\mapsto \wedge(TM)\otimes E$. The map $\nabla s$ takes $m\in M$ to the map $(\nabla s)_m: T_mM\rightarrow E_m$ (where $E_m$ is the fibre of $E$ at $m$), given by $(\nabla s)_m(X_m) := (\nabla_Xs)(m)$. Here $X\in\mathfrak{X}(M)$ is any vector field that agrees with $X_m$ at $m$. One can show from the properties of the connection that this quantity is independent of the choice of extension $X$.