Using some rather compact notation, we have that if $\mathfrak{e}=\mathfrak{v}R_\phi$, then $$\omega = R_{-\phi}\theta R_\phi + R_{-\phi} {\rm d}(R_\phi),$$because this is the general transformation law for local connection $1$-forms relative to two different frames, in terms of the change-of-basis matrix. Now, this becomes $$\begin{pmatrix} 0 & \omega_{12} \\ -\omega_{12} & 0\end{pmatrix} = \begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{pmatrix}\begin{pmatrix} 0 & \theta_{12} \\ -\theta_{12} & 0\end{pmatrix}\begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi\end{pmatrix}+ \begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi\end{pmatrix}\begin{pmatrix} -\sin\phi & -\cos\phi \\ \cos\phi & -\sin\phi\end{pmatrix}\,{\rm d}\phi. $$Looking only at the $(1,2)$-entry, the desired formula follows after applying $\cos^2\phi+\sin^2\phi=1$.
Proof of transformation law. If $(M,\nabla)$ is any affine manifold (i.e., $M$ is equipped with an arbitrary affine connection $\nabla$; it may have torsion, $M$ doesn't need to have a metric, we're doing it freestyle), and $\mathfrak{e}$ and $\mathfrak{v}$ are local frames with connection $1$-form matrices $\theta$ and $\omega$, and related via $\mathfrak{e} = \mathfrak{v}A$ for some non-singular matrix of functions $A$, we'll show that $$\theta = A^{-1}\omega A + A^{-1}{\rm d}A.$$
Namely, our assumptions fully written are that $$e_j = \sum_i a^i_{\,j}v_i,\quad \nabla e_j = \sum_i\theta^i_{\,j}\otimes e_i,\quad \nabla v_j = \sum_i \omega^i_{\,j}\otimes v_i.$$We'll write the latter two simply as $$\nabla\mathfrak{e} = \mathfrak{e}\theta\quad\mbox{and}\quad \nabla\mathfrak{v} = \mathfrak{v}\omega.$$Thus, we compute $$\mathfrak{v}A\theta = \mathfrak{e}\theta = \nabla\mathfrak{e} = \nabla(\mathfrak{v}A) = (\nabla\mathfrak{v})A+\mathfrak{v}\,{\rm d}A = \mathfrak{v}\omega A +\mathfrak{v}\,{\rm d}A = \mathfrak{v}(\omega A + {\rm d}A), $$so linear independence of $\mathfrak{v}$ gives $$A\theta = \omega A + {\rm d}A \implies \theta = A^{-1}\omega A + A^{-1}\,{\rm d}A.$$Same proof gives same transformation law for connection $1$-forms in an arbitrary vector bundle with connection, and two local trivializations.
You can try to read my short notes on Cartan computations, pages 11 through 14 of my notes on principal bundles, and the book by Loring Tu (Amazon link).
No, you can't determine the connection just from the torsion.
The space of linear connections on a given manifold $M$ is an affine space whose associated translation space is the space of type $(1,2)$-tensor fields on $M$. This means that given any two connections $\nabla,\nabla'$ on $M$, then $\nabla'-\nabla$ is a type $(1,2)$-tensor field, and if $\nabla$ is a connection and $A$ is a type $(1,2)$-tensor field, then $\nabla+A$ is a connection.
So, fix a connection $\nabla$ and a such a tensor $A$. Compute:
$$\begin{align}\tau^{\nabla+A}(X,Y) &= (\nabla+A)_XY - (\nabla+A)_YX - [X,Y] \\ &= \nabla_XY+A_XY-\nabla_YX-A_YX-[X,Y] \\ &= \tau^\nabla(X,Y) + A_XY-A_YX.\end{align}$$This means that $\nabla+A$ and $\nabla$ have the same torsion whenever $A$ is symmetric in the sense that $A_XY=A_YX$, for all vector fields $X$ and $Y$.
Best Answer
Note that $A$ is the matrix whose row vectors are the vectors $E_i$ written in terms of the standard basis $U_j$. So $$\omega_{ij}= \langle\nabla E_i,E_j\rangle = \langle dE_i,E_j\rangle = (dA\,A^\top)_{ij},$$ as required.