I have the following formula in 3-Dimensions:
$$ \int_{\partial \omega} f(x,y,z) \vec{dS} = \int_{\omega} \nabla f dV$$
I want to write the above in the language of forms and derive it through stokes.
My attempt:
Let $\tilde{R}(u,v)$ be the position vector on the surface, we have the two tangent vectors as $\tilde{R}_u $ and $\tilde{R}_v$. As a form , I write as $r_u$ and $r_v$. The integral is rewritten as refer :
$$ \int_{\partial w} f(x,y,z) \frac{ r_u \wedge r_v}{|r_u \wedge r_v|} = \int_{\partial \omega} f(x,y,z) n$$
Where $n$ is the normal two form.
By stokes theorem:
$$ \int_{\partial w} fn = \int_{\omega} d(fn) = \int_{\omega} df \wedge n + f dn $$
Not sure how to proceed now. I am suppose the exterior derivative of the normal relates to the Riemann curvature tensor which I am nto so familiar about yet.
Maybe this will be helpful
Classically we have two theorems in $\mathbb R^3$: Gauss' divergence theorem and Stokes' theorem. Your formula confuses me a bit as it seems to be neither one of them.
A better notation is probably $$\tag{1} \int_{\partial\omega}v=\int_\omega dv $$ where $$\tag{2} v=v_1\,(dy\wedge dz)+v_2\,(dx\wedge dz)+v_3\,(dx\wedge dy) $$ is a $2$-form. It looks like you are interested only in the cases $v=(f,0,0)$ or $v=(0,f,0)$ or $v=(0,0,f)$. (However I don't know what your notation $f\,\vec{dS}$ exactly means.)
The exterior derivative of $v$ is a $3$-form
\begin{align} dv&=(\partial_x v_1)\,(dx\wedge dy\wedge dz)-(\partial_y v_2)\,(dx\wedge dy\wedge dz)+ (\partial_z v_3)\,(dx\wedge dy\wedge dz)\\ &=\Big(\partial_x v_1-\partial_y v_2+\partial_z v_3\Big)(dx\wedge dy\wedge dz)\,. \end{align} Obviously, we now have a divergence: $$ \partial_x v_1-\partial_y v_2+\partial_z v_3=\nabla\begin{pmatrix}v_1\\-v_2\\v_3\end{pmatrix} $$ which gets integrated over the volume $\omega\,$. In other words, we have Gauss' theorem.
To cast Stokes' theorem into the form (1) note that this deals with a line integral over a closed curve $\gamma$ that is the boundary of a surface $\omega$. So here $v$ is a $1$-form
$$ v=v_1\,dx+v_2\,dy+v_3\,dz $$ and \begin{align} dv =&-(\partial_yv_3)\,(dy\wedge dz)+(\partial_z v_2)\,(dy\wedge dz)\\ &-(\partial_zv_1)\,(dx\wedge dz)+(\partial_x v_3)\,(dx\wedge dz)\\ &-(\partial_y v_1)\,(dx\wedge dy)+(\partial_x v_2)\,(dx\wedge dy). \end{align} Obviously, we now have a rotation $$ dv = (\nabla\times v)\cdot\begin{pmatrix}dy\wedge dz\\ dx\wedge dz\\ dx\wedge dy \end{pmatrix} $$ whose dot product gets integrated over the now two dimensional surface $\omega$.
The beauty of Elie Cartan's exterior calculus is that it automatically leads to the right expressions of divergence, resp. rotation.
peek-a-boo has kindly pointed out that your surface integral $\int_{\partial \omega}f\vec{dS}$ is that of a vector valued $2$-form whose components are those of $f$ multiplied with the unit outward normal vector $\vec{\boldsymbol{n}}$ to the surface. My favourite book on vector valued or tensor valued forms is Misner, Thorne and Wheeler Gravitation. I believe they would write this vector valued form as $$\tag{3} \boldsymbol{v}=\boldsymbol{v}_1\,(dy\wedge dz)+\boldsymbol{v}_2\,(dx\wedge dz)+\boldsymbol{v}_3\,(dx\wedge dy) $$ where each $\boldsymbol{v}_i$ is now a vector field in $\mathbb R^3$. In other words, each $\boldsymbol{v}_i$ has components $v_{ij}\,$: $$ \boldsymbol{v}_i=\begin{pmatrix}v_{i1}\\v_{i2}\\v_{i3}\end{pmatrix} $$ The form (3) is a stack of three forms of type (2) and clearly Gauss theorem yields a stack of three equations of type (1): $$ \int_{\partial \omega}\boldsymbol{v}=\int_\omega d\boldsymbol{v} $$ where the vector valued $3$-form has components
$$ d\boldsymbol{v}=\begin{pmatrix}\partial_x v_{11}-\partial_y v_{12}+\partial_z v_{13}\\\partial_x v_{21}-\partial_y v_{22}+\partial_z v_{23}\\\partial_x v_{31}-\partial_y v_{32}+\partial_z v_{33} \end{pmatrix}(dx\wedge dy\wedge dz)\,. $$ I.e., each component comes from the divergence of a vector field $$ \begin{pmatrix}v_{i1}\\-v_{i2}\\v_{i3}\end{pmatrix}\,. $$