For a general vector bundle, I there is no "$E$-valued integration" as you put it. You are trying to add up elements in the fibres of $E$, but since the fibres over different points are not the same vector space you can't add their elements.
For the trivial bundle $M \times \mathbb{R}^k$ - and with a fixed choice of trivialisation! - you can carry out the integral component by component, But if you change the trivialisation you will get a different answer. Moreover, you can change the trivialisation in a way that varies over the manifold, so there is no hope that the integral will just change by a linear map of $\mathbb{R}^k$. You can see this behaviour even with ordinary functions. A function is a section of the trivialised rank 1 bundle. If you change the trivialisation, but insist on regarding ordinray $p$-forms as bundle valued, you multiply all your forms by a fixed nowhere vanishing function. This can change the integral over a $p$-cycle in a more-or-less arbitrary way.
What you can integrate $E$-valued forms against is $E^*$-valued ones. Given $a \in \Omega^p(E)$ and $b \in \Omega^q(E^*)$ their wedge product is an ordinary $(p+q)$-form which you can then integrate over a $(p+q)$-cycle. Now you have a version of Stokes theorem. If you have a connection $A$ in $E$ then you can check that
$$
d(a \wedge b) = d_A(a) \wedge b \pm a \wedge d_A(b).
$$
So Stokes theorem gives
$$
\int_{M}d_A(a) \wedge b \pm a \wedge d_A(b) = \int_{\partial M} a \wedge b.
$$
In the case when $b$ is a covariant constant section of $E$ and $M$ has dimension one more than the degree of $a$, we get
$$\int_M \langle d_A(a) , b \rangle=\int_{\partial M} \langle a, b \rangle$$
This is just the usual Stokes theorem, for the component of $a$ in the direction $b$. Since $b$ is covariant-constant, $\langle d_A(a), b \rangle = d \langle a, b \rangle$.
The question you are asking is a very basic one in the theory of what Élie Cartan called "the method of the moving frame" (in the original French, "la méthode du repère mobile"), so you should be looking that up. Cartan's basic goal was to understand maps of manifolds into homogeneous spaces, say, $f:M\to G/H$, by associating to each such $f$, in a canonical way, a 'lifting' $F:M\to G$ in such a way that the lifting of $\hat f = g\cdot f$ would be $\hat F = gF$ for all $g\in G$. If one could do such a thing, then one could tell whether two maps $f_1,f_2:M\to G/H$ differed by an action of $G$ by checking whether $F_1^*(\gamma) = F_2^*(\gamma)$, where $\gamma$ is the canonical $\frak{g}$-valued left-invariant $1$-form on $G$.
It turns out that it is not always possible to do this in a uniform way for all smooth maps $f:M\to G/H$ (even in the pointed category, which modifies the problem a little bit, but not by much). However, if one restricts attention to the maps satisfying some appropriate open, generic conditions, then there often is a canonical lifting $F$ for those $f$ belonging to this set of mappings, and it can be characterized exactly by requiring that the $1$-form $\omega_F = F^*(\gamma)$ satisfy some conditions. Working out these conditions in specific cases is what is known as the "method of the moving frame".
There's no point in trying to give an exposition of the theory here because it is covered in many texts and articles, but let me just give one specific example that should be very familiar, the Frenet frame for Euclidean space curves.
Here the group $G$ is the group of Euclidean motions (translations and rotations) of $\mathbb{E}^3$ and $H$ is the subgroup that fixes the origin $0\in\mathbb{E}^3$. The elements of $G$ can be thought of as quadruples $(x,e_1,e_2,e_3)$ where $x\in\mathbb{E}^3$ and $e_1,e_2,e_3$ are an orthonormal basis of $\mathbb{E}^3$.
When $f:\mathbb{R}\to\mathbb{E^3}$ is nondegenerate, i.e., $f'(t)\wedge f''(t)$ is nonvanishing, there is a canonical lifting $F:\mathbb{R}\to G$ given by
$$
F(t) = \bigl(f(t),e_1(t),e_2(t),e_3(t)\bigr)
$$
that is characterized by conditions on $\omega = F^*(\gamma)$ that are phrased as follows: First, $e_1\cdot df$ is a positive $1$-form
while $e_2\cdot df = e_3\cdot df = 0$, and, second, $e_2\cdot de_1$ is a positive $1$-form while $e_3\cdot de_1 = 0$.
These conditions take the more familiar form
$$
df = e_1(t)\ v(t)dt,\qquad
de_1 = e_2(t)\ \kappa(t)v(t)dt,
$$
for some positive functions $v$ and $\kappa$ on $\mathbb{R}$, and they imply
$$
de_2 = -e_1(t)\ \kappa(t)v(t)dt + e_3(t)\ \tau(t)v(t)dt,
\qquad
de_3 = -e_2(t)\ \tau(t)v(t)dt,
$$
for some third function $\tau$ on $\mathbb{R}$.
Conversely, any $F:\mathbb{R}\to G$ that satisfies the above conditions on $\omega = F^*(\gamma)$ is the canonical (Frenet) lift of a (unique) nondegenerate $f:\mathbb{R}\to\mathbb{E}^3$.
Without the nondegeneracy condition, the uniqueness fails. Just consider the case in which the image of $f$ is a straight line.
There are similar, but, of course, more elaborate, examples for other homogeneous spaces and higher dimensional $M$, but you should go look at the literature if you are interested in this.
Added in response to request in the comment: There are several excellent sources for the method of the moving frame. I'll just list (alphabetically by author) some of my favorites, which means the ones that I find most felicitous:
Élie Cartan, "La théorie des groupes finis et continus et la géométrie différentielle traitées par la méthode du repère mobile", Paris: Gauthier-Villars, 1937. (His style takes some getting used to, and so many say that Cartan is unreadable, but, once you get used to the way he writes, there's nothing like Cartan for clarity and concision. I certainly have learned more from reading Cartan than from any other source.)
Shiing-shen Chern, W. H. Chen, and K. S. Lam, "Lectures on Differential Geometry", Series on University Mathematics, World Scientific Publishing Company, 1999. (Chern learned from Cartan himself, and was a master at calculation using the method.)
Jeanne Clelland, 1999 MSRI lectures on Lie groups and the method of moving frames, available at http://math.colorado.edu/~jnc/MSRI.html. (A nice, short elementary introduction.)
Mark Green, "The moving frame, differential invariants and rigidity theorems for curves in homogeneous spaces", Duke Math. J. Volume 45, Number 4 (1978), 735-779. (Points out some of the subtleties in the 'method' and that it sometimes has to be supplemented with other techniques.)
Phillip Griffiths, " "On Cartan’s method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry", Duke Math. J. 41 (1974): 775–814. (Lots of good applications and calculations.)
Thomas Ivey and J. M. Landsberg, "Cartan for Beginners: Differential Geometry Via Moving Frames and Exterior Differential Systems", Graduate Studies in Mathematics, AMS, 2003. (Also contains related material on how to solve the various PDE problems that show up in the applications of moving frames.)
I think that this is enough to go on. I won't try to go into some modern aspects, such as the work of Peter Olver and his coworkers and students, who have had some success in turning Cartan's method into an algorithm under certain circumstances, or the more recent work of Boris Doubrov and his coworkers on applying Tanaka-type ideas to produce new approaches to the moving frame in certain cases.
Best Answer
Your confusion is revealed in this sentence "Or one defines something called the exterior covariant derivative D (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form." This is just not true; one does not take the `exterior covariant derivative of the connection $1$-form' to get the curvature.
Let's be precise: Let $P\to M$ be a principal right $G$-bundle, and let $\omega$ be a $\frak{g}$-valued $1$-form on $P$ that defines a connection on $P$ (I won't repeat the well-known requirements on $\omega$). The curvature $2$-form $\Omega = d\omega +\frac12[\omega,\omega]$ on $P$ is the $2$-form that vanishes if and only if it is possible to find local trivializations $\tau: P_U \to U\times G$ such that $\omega = (\pi_2\circ\tau)^*(\gamma)$ where $\gamma$ is the canonical left-invariant $1$-form on $G$.
Now, given a representation $\rho:G\to \text{GL}(V)$, where $V$ is a vector space, one can define an associated vector bundle $E = P\times_\rho V$. Using $\omega$, it is possible to define an 'exterior covariant derivative operator' $D_\omega:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+1})$ (where $A^p\to M$ is the bundle of alternating (i.e., 'exterior') $p$-forms on $M$). The operator ${D_\omega}^2:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+2})$ then turns out to be linear over the $C^\infty$ alternating forms, so it is determined by its value when $p=0$, i.e., by ${D_\omega}^2:\Gamma(E)\to \Gamma(E\otimes A^{2})$, which can be regarded as an section of $\text{End}(E)\otimes A^2$, i.e., a $2$-form with values in $\text{End}(E)$. The formula for this section, when pulled back to $P$, can now be expressed in terms of $\Omega$ in the usual way. In particular, ${D_\omega}^2$ vanishes identically if $\Omega$ does.
Note that one does not take the 'exterior covariant derivative' of the $1$-form $\omega$ anywhere. Instead, one takes the exterior covariant derivative of the exterior covariant derivative of a section of $E$.
I suspect that what you may be trying to do is interpret $\omega$ as a $1$-form on $P$ with values in the trivial bundle $P\times\frak{g}$ and then say that the curvature is the 'exterior covariant derivative' of $\omega$. However, to make this work, you have to specify a connection on the trivial bundle $P\times\frak{g}$, which amounts to choosing a $1$-form $\eta$ on $P$ that takes values in $\text{End}(\frak{g})$ and setting $D_\eta(s) = ds + \eta\ s$. By setting $\eta = \frac12\text{ad}(\omega)$, one gets $D_\eta\omega = \Omega$ (just by definition), so it is possible to do this, but I don't think that this is that useful an observation, since, after all, you could have taken the connection on the trivial bundle $P\times\frak{g}$ to be $\eta = \frac13\text{ad}(\omega)$ (for example) or even $\eta=0$. What justifies the $\frac12$, other than the desire to get the `right' answer?