Dual frame in Riemannian metrics.

I think you can do this by direct calculation. Fix $p\in U$. If $X_p\in TU_p$, then $X_p=a\partial_u+b\partial_v$ for some $a,b\in \mathbb R$, and by definition of the $1-$forms $du$ and $dv$, we have $du(X_p)=a$ and $dv(X_p)=b$. Now, apply this to the tangent vectors $e_1$ and $e_2$:

$\omega_1(e_1)=\sqrt{E}\left(du+\frac{F}{E}dv\right)\frac{1}{\sqrt{E}}\frac{\partial}{\partial u}=\sqrt E\frac{1}{\sqrt E}(\partial_uu)=\frac{\sqrt E}{\sqrt E}=1$

Similarly,

$\omega_1(e_2)=\sqrt{E}\left(du+\frac{F}{E}dv\right)\left(\frac{-1}{\sqrt{EG-F^2}}\left(\frac{F}{\sqrt{E}}\frac{\partial}{\partial u}-\sqrt{E}\frac{\partial}{\partial v}\right)\right )=-\frac{\sqrt E}{\sqrt{EG-F^2}}\left (\frac{F}{\sqrt E}-\frac{F\sqrt E}{E}\right )=0.$

$\omega_2(e_1)=\sqrt{\frac{EG-F^2}{E}}dv\left(\frac{1}{\sqrt{E}}\frac{\partial}{\partial u} \right)=\sqrt{\frac{EG-F^2}{E^2}}\partial_u v=0$

and

$\omega_2(e_2)=\sqrt{\frac{EG-F^2}{E}}dv\left(\frac{-1}{\sqrt{EG-F^2}}\left(\frac{F}{\sqrt{E}}\frac{\partial}{\partial u}-\sqrt{E}\frac{\partial}{\partial v}\right)\right)=\left(\sqrt{\frac{EG-F^2}{E}}\right)\cdot \left(\frac{\sqrt E}{\sqrt{EG-F^2}}\right)=1$

Note that $e_i$ is an orthonormal vector fields wrt $ds^2$. Hence we let $ \omega_i=ds^2(e_i,\cdot)$ which is dual frame.