On any trivial vector bundle $E=M\times V$ (with $M$ a smooth manifold and $V$ a finite-dimensional real/complex vector space), the trivial connection on $E$ is defined such that for all sections $\psi$ of $E$, we have $\nabla\psi=d\psi$, the latter being the exterior derivative of a vector-valued map (which you can define using any basis). More explicitly, $\psi$ being a section means it is of the form $\psi(x)=(x,\Psi(x))$ for some smooth map $\Psi:M\to V$. So, for any vector field $X$ on $M$, we define $(\nabla_X\psi)(x):=(x,(d\Psi)_x(X_x))$. This is equivalent to the following: for any basis $\{v_1,\dots, v_k\}$ of $V$, let $\{e_1,\dots, e_n\}$ be the corresponding sections (i.e $e_i(x)=(x,v_i)$); then we require $\nabla(e_i)=0$ for all $i\in\{1,\dots, k\}$.
So, in general if you have a section $\psi$, and you expand it as $\psi=\sum_{i=1}^k\psi^i\,e_i$ for some smooth functions $\psi_i$, then $\nabla\psi=\sum_{i=1}^k(d\psi^i)\,e_i$, i.e $\nabla_X\psi=\sum_{i=1}^k(d\psi^i)(X)\cdot e_i=\sum_{i=1}^kX(\psi^i)\cdot e_i$, where the latter is the action of a vector field on a smooth function, etc. Another way to express this condition is that the connection $1$-forms relative to the frame $\{e_1,\dots, e_k\}$ (recall, these are the unique $1$-forms $\omega^{i}_{\,j}$ on $M$ such that $\nabla(e_j)=\sum_{i=1}^k\omega^{i}_{\,j}\otimes e_i$; see here for more elaboration (also what I say there extends to general vector bundles)) vanish. The point is there are many ways to express this condition, and the reason it’s called the trivial connection is because if you think for a moment, the definition says that you actually don’t need to know anything about vector bundles and connections and so on; this is just the usual directional derivative you learn about in your first introduction to multivariable calculus.
As far as generalities go, my answer to Curvature of a connection $\nabla + A$ on a vector bundle $E$. addresses your situation fully as well. But, I realize it may not be the easiest to digest on first glance, so let’s do this in a more down-to-earth (but also less ‘efficient’) manner.
Let’s start with a general vector bundle $(E,\pi,M)$ with a connection $\nabla$, and let’s fix a local-trivialization over $U\subset M$ for it; this is equivalent to specifying a local frame $\{e_1,\dots, e_k\}$ for $E$ over $U$.
Let $\omega^{i}_{\,j}$ be the connection $1$-forms of $\nabla$ relative to this local frame, as we have defined above. So, for any section $\psi$, if we write it relative to the local frame as $\psi=\psi^i\,e_i$, then by the product rule for connections,
\begin{align}
\nabla\psi&=(d\psi^i)\otimes e_i+\psi^j\nabla(e_j)\\
&=(d\psi^i)\otimes e_i+\psi^j\cdot\omega^i_{\,j}\otimes e_i\\
&=\left(d\psi^i+\omega^{i}_{\,j}\psi^j\right)\otimes e_i
\end{align}
This equation is often abbreviated as $\nabla=d+\omega$ or $\nabla=\nabla^{\text{trivial}}+\omega$.
Note also that for each choice of vector fields $X,Y$ on $U$, the curvature $R(X,Y)$ is an endomorphism of the various fibers of $E$: it eats a section $\psi$ and outputs a section $R(X,Y)\cdot\psi$. So, for each $p\in U$, $R(X,Y)_p:E_p\to E_p$ is a linear map, and so relative to the basis $\{e_1(p),\dots, e_k(p)\}$ of $E_p$, this has a $k\times k$ matrix representation. Omitting the point $p$, we denote this as $\Omega^{i}_{\,j}(X,Y)$; more succinctly, $\Omega^{i}_{\,j}(\cdot,\cdot)=[R(\cdot,\cdot)]^i_{\,j}$. Also, note that this is $C^{\infty}$-bilinear and alternating in $X,Y$ since $R$ is. Hence, for each choice of index $i,j\in\{1,\dots, k\}$, the object $\Omega^{i}_{\,j}$ is a $2$-form on $U$, called the $(i,j)$ curvature 2-form of $\nabla$ relative to the local frame $\{e_1,\dots, e_k\}$, or simply the curvature $2$-forms for short.
So, notice that knowledge of all the $2$-forms $\Omega^{i}_{\,j}$ is equivalent to knowledge of the entire curvature $R$. Now, Cartan’s second structure equations relative the $\Omega^i_{\,j}$ to the $\omega^i_{\,j}$’s:
\begin{align}
\Omega^{i}_{\,j}&=d\omega^i_{\,j}+\sum_{a=1}^k\omega^i_{\,a}\wedge\omega^a_{\,j},
\end{align}
or simply $\Omega=d\omega+\omega\wedge\omega$. See planetmath for a tedious but understandable proof (just replace the $\partial_i$ there with $e_i$ so that it works in our vector bundle case)).
Your Example.
Henceforth I will suppress the indices and write everything as matrices. In your case, we shall take the standard basis of the vector space $\Bbb{R}^2$ and upgrade them to sections of the trivial vector bundle $\{e_1,e_2\}$ as I described in my first paragraph. Then, we have
\begin{align}
\omega&=A\,dx+B\,dy=
\begin{pmatrix}
y\,dx-x\,dy&dx\\
0&y\,dx-x\,dy
\end{pmatrix}.
\end{align}
So, by Cartan’s second structure equation, the connection $2$-forms are given by
\begin{align}
\Omega&=d\omega+\omega\wedge\omega\\
&=
\begin{pmatrix}
-2\,dx\wedge dy&0\\
0&-2\,dx\wedge dy
\end{pmatrix}
+
\begin{pmatrix}
0&0\\
0&0
\end{pmatrix}\\
&=
\begin{pmatrix}
-2\,dx\wedge dy&0\\
0& -2\,dx\wedge dy
\end{pmatrix},
\end{align}
where I should warn you that in the penultimate step, $\omega\wedge\omega$ is not zero in general simply by ‘anticommutativity of wedge product of $1$-forms’, because matrix multiplication is not commutative. However, in this special case, the wedge product of the matrices is $0$.
