Consider a smooth surface $(\Sigma \subset R^{3}, \phi)$, where $\phi(x,y):\, D\subset R^{2} \mapsto R^{3}$ is a parametrization of $\Sigma$, $D$ is a open set in $R^{2}$.
As we know that at every point $(x,y, \phi(x,y))$ on $\Sigma$ which has a tangent space spanned by $\lbrace \phi_{x}, \, \phi_{y} \rbrace$ and the dual basis $\lbrace dx, \, dy \rbrace$. Then, when we are considering a $1-$ form on $\Sigma$ having a parametrization $\omega=a(x,y)dx+b(x,y)dy$, we have $$d\omega=(b_{x}-a_{y})dx \wedge dy.$$ Now the question is:
If $\omega=c(x,y)e_{1}+d(x,y)e_{2}$, where $e_{1}, e_{2}$ are the orthogonal frame after using gram schmidt orthogonalization. How can I compute $d\omega$ in this frame?.
Please don't use $e_i$ for differential forms. You should write an orthonormal basis for the tangent space as $\{e_1,e_2\}$ and let $\omega_1,\omega_2$ be the dual basis for the cotangent space. The connection form $\omega_{12}$ is uniquely determined by the structure equations $$d\omega_1=\omega_{12}\wedge\omega_2 \quad\text{and}\quad d\omega_2=-\omega_{12}\wedge\omega_1.$$ Then, if $\phi = f\omega_1+g\omega_2$, you differentiate, as usual, getting \begin{align*} d\phi &= df\wedge\omega_1 + f(\omega_{12}\wedge\omega_2) + dg\wedge\omega_2 + g(-\omega_{12}\wedge\omega_1 \\ &= (df-g\omega_{12})\wedge\omega_1 + (dg+f\omega_{12})\wedge\omega_2. \end{align*} To calculate the connection form, of course, you need to know the induced metric on your surface and actually compute the coframe $\omega_1,\omega_2$ explicitly. (Many times it is better to find an orthogonal parametrization, rather than a graph parametrization. Of course, sometimes that's hard to do in practice.)