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?
First, there is a canonical isomorphism $\Phi$ between $\Omega^k(M; Ad P)$ and $\Omega^k_{Ad, h}(P; \mathfrak{g})$, where the subscripts signify that the form is horizontal and of type $Ad$ (i.e. equivariant). This isomorphism is canonical and does not need a local trivialization. To illustrate the idea, consider the case $k=0$. A section $\varphi$ of the adjoint bundle defines a function $\Phi(\varphi)$ on $P$ with values in $\mathfrak{g}$ by $\varphi(m) = [p, \Phi(\varphi) (p)]$ where $\pi(p) = m$. In your notation, $\Phi^{-1}(F) = \mathcal{F}$.
Next, the exterior differential in the adjoint bundle (indeed in every associated bundle) is defined by making the following diagram commutative:
$$\begin{matrix} \Omega^k_{Ad, h}(P; \mathfrak{g}) & \overset{D}{\to} & \Omega^{k+1}_{Ad, h}(P; \mathfrak{g}) \\ \downarrow & & \downarrow \\ \Omega^k(M; Ad P) & \overset{d_{\nabla}}{\to} & \Omega^{k+1}(M; Ad P) \end{matrix}$$
(the vertical arrows being the isomorphism $\Phi$)
Thus $D F = 0$ coincides with $d_{\nabla}\mathcal{F} = 0$ under the isomorphism $\Phi$.
Finally, there is also the viewpoint via local trivializations. Locally, we can identify the connection form with a Lie algebra valued 1-form $\tilde{A}$ on $M$ and the curvature is a 2-form $\tilde{F}$ on $M$ with values in the Lie algebra. It is an easy calculation to see that $D$ locally takes the form $\widetilde{D F} = d \tilde{F} + [\tilde{A}, \tilde{F}]$, i.e. the local representant of $DF$ is given by the right-hand side. Note that this expression only makes sense locally, since the differential of $\mathcal{F}$ as a vector-valued 2-form is not well-defined (without a connection). Nonetheless, $d_\nabla$ is always defined and can be applied to $\mathcal{F}$ without any problems.
Best Answer
Your definition is a sort of hybrid of two standard definitions of connection 1-form. These are:
There is a one-to-one correspondence between connection 1-forms and Ehresmann connections as follows. Given a connection 1-form $\omega$, the kernel of $\omega$ is $G$-invariant and complementary to $VP$, so it defines an Ehresmann connection. Conversely, given an Ehresmann connection $TP = HP \oplus VP$, let $q: TP \to VP$ denote the natural projection map (a morphism of vector bundles). Note that the vector fields $A^\#$ define an isomorphism between $VP$ and the trivial bundle $P \times \mathfrak{g} \to P$, so composing $q$ with the projection $VP \cong P \times \mathfrak{g} \to \mathfrak{g}$ gives a map $TP \to \mathfrak{g}$, i.e. a $\mathfrak{g}$-valued 1-form. This is a connection 1-form.
Unwinding these identifications, you could just as well use a third definition:
This is your definition. From here you can deduce that $H$ complements $VP$ in $TP$, (or that $\omega$ is compatible with the adjoint action of $G$) and hence connections in this sense are the same as the objects defined in the previous two definitions.