[proofplan]
We first compute the Levi-Civita connection matrix in the oriented orthonormal frame. Metric compatibility forces the matrix to be skew-symmetric, and the chosen convention $\nabla_X e_1=\alpha(X)e_2$ fixes the signs of its entries. Since $\mathfrak{so}(2)$ is abelian, the curvature matrix has no quadratic bracket contribution and is obtained by applying $d$ to the scalar coefficient. Finally, comparing the curvature matrix with the Riemann curvature operator on the ordered frame $(e_1,e_2)$ identifies $d\alpha(e_1,e_2)$ with $-K$, which is exactly the displayed formula because $\theta_1\wedge\theta_2$ is the oriented area form of the frame.
[/proofplan]
[step:Write the connection matrix in the oriented orthonormal frame]
Let $\mathfrak{X}(U)$ denote the space of smooth vector fields on $U$, and let $X \in \mathfrak{X}(U)$. By definition of $\alpha$,
\begin{align*}
\nabla_X e_1 = \alpha(X)e_2.
\end{align*}
Since $(e_1,e_2)$ is $g$-orthonormal and $\nabla$ is metric-compatible,
\begin{align*}
0 = X(g(e_1,e_2)) = g(\nabla_X e_1,e_2)+g(e_1,\nabla_X e_2).
\end{align*}
Substituting $\nabla_X e_1=\alpha(X)e_2$ gives
\begin{align*}
g(e_1,\nabla_X e_2) = -\alpha(X).
\end{align*}
Also,
\begin{align*}
0 = X(g(e_2,e_2)) = 2g(\nabla_X e_2,e_2),
\end{align*}
so $\nabla_X e_2$ has no $e_2$ component. Therefore
\begin{align*}
\nabla_X e_2 = -\alpha(X)e_1.
\end{align*}
With the convention $\nabla_X e_j=\sum_{i=1}^2 A_{ij}(X)e_i$, the connection matrix is determined by the four entries
\begin{align*}
A_{11}=0, \quad A_{12}=-\alpha, \quad A_{21}=\alpha, \quad A_{22}=0.
\end{align*}
[guided]
We must determine all four entries of the local connection matrix, not only the one encoded by the definition of $\alpha$. The convention is column-based: the $j$-th column records the coefficients of $\nabla_X e_j$ in the basis $(e_1,e_2)$.
The definition of the connection form gives the first column immediately:
\begin{align*}
\nabla_X e_1 = 0\cdot e_1+\alpha(X)e_2.
\end{align*}
Thus $A_{11}=0$ and $A_{21}=\alpha$.
Now we compute the second column from metric compatibility. Since $e_1$ and $e_2$ are orthonormal, $g(e_1,e_2)=0$ and $g(e_2,e_2)=1$ on $U$. Applying $X$ to these identities and using $\nabla g=0$ gives
\begin{align*}
0 = X(g(e_1,e_2)) = g(\nabla_X e_1,e_2)+g(e_1,\nabla_X e_2)
\end{align*}
and
\begin{align*}
0 = X(g(e_2,e_2)) = 2g(\nabla_X e_2,e_2).
\end{align*}
The second identity says that $\nabla_X e_2$ has no $e_2$ component. The first identity, together with $\nabla_X e_1=\alpha(X)e_2$, gives
\begin{align*}
0 = \alpha(X)g(e_2,e_2)+g(e_1,\nabla_X e_2),
\end{align*}
hence
\begin{align*}
g(e_1,\nabla_X e_2) = -\alpha(X).
\end{align*}
Therefore
\begin{align*}
\nabla_X e_2 = -\alpha(X)e_1.
\end{align*}
So the matrix of $1$-forms satisfying $\nabla_X e_j=\sum_{i=1}^2 A_{ij}(X)e_i$ is determined by
\begin{align*}
A_{11}=0, \quad A_{12}=-\alpha, \quad A_{21}=\alpha, \quad A_{22}=0.
\end{align*}
[/guided]
[/step]
[step:Compute the curvature matrix using the abelian Lie algebra $\mathfrak{so}(2)$]
Let $\Omega \in \Omega^2(U;\mathfrak{so}(2))$ denote the curvature matrix of the connection $A$. In a local frame,
\begin{align*}
\Omega = dA + A\wedge A,
\end{align*}
where $(A\wedge A)_{ij}=\sum_{k=1}^2 A_{ik}\wedge A_{kj}$. Since every entry of $A\wedge A$ is a wedge product of $\alpha$ with itself or with $0$, and $\alpha\wedge\alpha=0$, we have
\begin{align*}
A\wedge A = 0.
\end{align*}
Thus $\Omega=dA$ has entries
\begin{align*}
\Omega_{11}=0, \quad \Omega_{12}=-d\alpha, \quad \Omega_{21}=d\alpha, \quad \Omega_{22}=0.
\end{align*}
Equivalently, let $J \in \mathfrak{so}(2)$ denote the standard generator defined by its entries $J_{11}=0$, $J_{12}=-1$, $J_{21}=1$, and $J_{22}=0$. The curvature is
\begin{align*}
\Omega = d\alpha \, J.
\end{align*}
Hence the scalar local representative of the principal $SO(2)$ curvature is
\begin{align*}
F=d\alpha.
\end{align*}
[/step]
[step:Compare the curvature matrix with the Riemann curvature operator]
By definition of the curvature matrix, for smooth vector fields $X,Y \in \mathfrak{X}(U)$ and for $j\in\{1,2\}$,
\begin{align*}
R(X,Y)e_j = \sum_{i=1}^2 \Omega_{ij}(X,Y)e_i.
\end{align*}
Taking $j=2$ and using $\Omega_{12}=-d\alpha$ and $\Omega_{22}=0$, we get
\begin{align*}
R(X,Y)e_2 = -d\alpha(X,Y)e_1.
\end{align*}
Now evaluate at $X=e_1$ and $Y=e_2$. The definition of $K$ gives
\begin{align*}
K = g(R(e_1,e_2)e_2,e_1).
\end{align*}
Using the previous identity,
\begin{align*}
K = g(-d\alpha(e_1,e_2)e_1,e_1).
\end{align*}
Since $g(e_1,e_1)=1$,
\begin{align*}
K = -d\alpha(e_1,e_2).
\end{align*}
Thus
\begin{align*}
d\alpha(e_1,e_2) = -K.
\end{align*}
[/step]
[step:Identify the two-form from its coefficient on the oriented coframe]
Because $\dim \Sigma=2$, every $2$-form on $U$ is a smooth scalar multiple of $\theta_1\wedge\theta_2$. Let $f:U\to\mathbb{R}$ be the unique smooth function such that
\begin{align*}
d\alpha = f\,\theta_1\wedge\theta_2.
\end{align*}
Evaluating both sides on the ordered frame $(e_1,e_2)$ gives
\begin{align*}
d\alpha(e_1,e_2)=f(\theta_1\wedge\theta_2)(e_1,e_2).
\end{align*}
Since $\theta_i(e_j)=\delta_{ij}$,
\begin{align*}
(\theta_1\wedge\theta_2)(e_1,e_2)=1.
\end{align*}
Therefore
\begin{align*}
f=d\alpha(e_1,e_2)=-K.
\end{align*}
Substituting this coefficient into the expansion of $d\alpha$ yields
\begin{align*}
d\alpha = -K\,\theta_1\wedge\theta_2.
\end{align*}
This proves both the local curvature representation $F=d\alpha$ and the stated relation with Gaussian curvature.
[/step]
