Why are these two definitions of a connection equivalent?

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I'm working through Kai Köhler's "Differentialgeometrie und homogene Räume" (Differential geometry and homogenous spaces) and struggle to understand the notion of a connection on a vector bundle. In particular, I do not completely understand why two definitions are equivalent. The main definition that the author gives is the following:

Let $ E \to M$ be a vector bundle. A (covariant) connection $\nabla$ on $E$ is a $\mathbb{R}$-linear map $\nabla:\Gamma(M,E)\to \Gamma(M,T^*M \otimes E)$, that satisfies Leibniz's rule: $\forall f\in C^\infty(M),s\in \Gamma(M,E): \nabla(f\cdot s) = df \otimes s + f\nabla s$.

Apparently, it is equivalent to ask for $\nabla$ to be a map $\nabla:\Gamma(M,TM) \times \Gamma(M,E) \to \Gamma(M,E)$, that is $C^{\infty}$-linear in the first and satisfies the Leibniz rule in the second argument. Why does this hold? Also, what is a good intuition to think about $df \otimes s$ ?

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Let $X \in \Gamma(M, TM)$ and $s \in \Gamma(M, E)$.

If $\nabla$ is a connection according to the first definition, then we can produce $D$, a connection according to the second condition, by defining $D(X, s) := (\nabla s)(X)$ which is usually denoted $\nabla_Xs$.

Conversely, if $D$ is a connection according to the second definition, then we can produce $\nabla$, a connection according to the first condition, by defining $\nabla s := D(\cdot, s)$.


As for $df\otimes s$, note that $df \in \Gamma(M, T^*M)$ and $s \in \Gamma(M, E)$, so $df\otimes s \in \Gamma(M, T^*M\otimes E)$. Given a vector field $X \in \Gamma(M, TM)$, then $(df\otimes s)(X) = df(X)\otimes s = X(f)\otimes s = X(f)s$.