I came across this following way of defining connection and curvature which is not so obviously equivalent to the definitions as familiar in Riemannian Geometry books like say by Jost.


If $E$ is a complex vector bundle over a manifold $M$ then one defines the space of vector valued $p$-differential forms on them as $\Omega^p(M,E) = \Gamma ( \wedge ^p (T^\*M) \otimes E) $


The connection be defined as the map , $\nabla: \Gamma(E) \rightarrow \Omega^1(M,E)$ satisfying $\nabla(fX) = (df)X+f\nabla X$ where $f \in C^{\infty}(M)$ and $X \in \Gamma(E)$


{It might help if someone can make explicit as to what exactly is $(df)X$"?
To make sense this has to be an element of $\Omega^1(M,E)$}


Now one defines curvature of the connection as the map, $$R = \nabla \circ \nabla : \Gamma(E) \rightarrow \Omega^2(M,E)$$


Now it is not clear to me as to how this makes $ R \in \Omega^2(M,End(E))$ ?


And how does this match up with the more familiar form as,


$R(X,Y) = \nabla \_X \nabla \_Y - \nabla \_Y \nabla \_X - \nabla \_{[X,Y]}$


One can extend the definition of connection to be a map $\nabla: \Omega^p(M,E) \rightarrow \Omega^{p+1}(M,E)$
such that for $\omega \in \Omega^p(M)$ and $X \in \Gamma(E)$ one has,


$$\nabla(\omega X) = (d\omega)X + (-1)^{deg \omega}\omega \wedge \nabla X$$


{Here again it would be helpful if someone can can explain what is the meaning of $(d\omega)X$"?
 To make sense it has to be an element of $\Omega^{p+1}(M,E)$}


This looks very much like the de-Rham derivative of vector valued differential forms but the notation looks uneasy.



Definition of the curvature tensor 
You need to write $\omega X$ as $\omega \otimes X$. 


If you prove that $R: \Gamma(E) \to \Omega^2(M,E)$ is tensorial - i.e. $R(X,Y)(fS) = fR(X,Y)S$ where $S\in\Gamma(E)$, $f\in\mathcal{C}^\infty(M)$ and $X,Y\in TM$, then it follows that this mapping is indeed an element of ${\Omega} ^2(M,End(TM))$, because then the value of $R(X,Y)S$ at a point $m\in M$ depends only on $X\_m,Y\_m$ and $S\_m$ and not on their derivatives (even though we have defined $R$ via differentiation). The desired identification then follows from linear algebra $Hom(\Lambda^2 V \otimes E,E) \simeq Hom(\Lambda^2 V, \otimes E^\* \otimes E)$ & $End(E) \simeq E^\*\otimes E$.


The classical formula for curvature follows directly from the definition of the action of $\nabla$ on $\Omega^p(M,E)$.


