First of all, you need to understand what $dS$ is.
$dS$ is the form $dV$ contracted by the vector field $\nu$ on $\partial \Omega$. In other words, to find $dS(v_2,v_3,...,v_{n})$ for tangent vectors $v_2,v_3,...,v_{n}$ to $\partial \Omega$, you just evaluate $dV(\nu, v_2,v_3,...,v_{n})$. The motivation for this definition is that the area of a parallelogram $P$ is the same as the volume of the parallelepiped with base $P$ and height $1$.
So now note that,
$
\frac{\partial f}{\partial x_1} dV = d(f dx_2 \wedge dx_3 \wedge ...\wedge dx_n)
$
Letting $ \widehat{dx_1} = dx_2 \wedge dx_3 \wedge ...\wedge dx_n$
Let $p \in \partial \Omega$ and $v_2,v_3,...,v_n \in T_p( \partial \Omega)$ then
$
\begin{align*}
f(p) \widehat{dx_1}(v_2,v_3,...,v_n) &= f(p) dV(e_1,v_2,...,v_n)\\
&=f(p) dV(\textrm{proj}_{\nu(p)}(e_1),v_2,...,v_n)\\
&=f(p) dV(\nu_1(p)\nu(p),v_2,...,v_n)\\
&=f(p)\nu_1(p) dV(\nu(p),v_2,...,v_n)\\
&=f(p)\nu_1(p) dS(v_2,...,v_n)
\end{align*}
$
so as forms on $\partial S$, we have $f\widehat{dx_1} = f\nu_1dS$. The big step is the second line in the equality above, which is justified by writing $e_1$ as a linear combination of $\nu$ and $v_i$'s. All of the $v_i$ terms die because of the alternating property of forms.
Now we conclude that
$\begin{align*}
\int_\Omega \frac{\partial f}{\partial x_1} dV &= \int_\Omega d(f \widehat{dx_i})\\
&=\int_{\partial \Omega} f \widehat{dx_i} \text{by Stokes' theorem}\\
&=\int_{\partial \Omega} f \nu_1 dS
\end{align*}
$
If any of this does not make sense, please ask for follow up in the comments. Notation in differential geometry is horrible, and everyone makes it up for themselves I think. So I understand if some part of this is hard to parse. I am also no differential geometer, but I do a lot of calculus in high dimensional spaces (I work in several complex variables). This kind of thing used to frustrate me a lot.
Best Answer
There is a simple proof of Gauss-Green theorem if one begins with the assumption of Divergence theorem, which is familiar from vector calculus, \begin{equation} \int_{U}\mathrm{div}\,\mathbf{w}\,dx = \int_{\partial U} \mathbf{w}\cdot\mathbf{\nu}\,dS, \end{equation} where $\mathbf{w}$ is any $C^\infty$ vector field on $U\in\Bbb{R}^n$ and $\mathbf{\nu}$ is the outward normal on $\partial U$.
Now, given the scalar function $u$ on the open set $U$, we can construct the vector field \begin{equation} \mathbf{w}=(0,\ldots,0,u,0,\ldots,0), \end{equation} where $u$ is the $i$th component. Then, following the Divergence theorem, we have \begin{equation} \int_U \mathrm{div}\,\mathbf{w}\,dx=\int_U u_{x_i}\,dx =\int_{\partial U}\mathbf{w}\cdot\mathbf{\nu}\,dS =\int_{\partial U}u\nu^i\,dS. \end{equation}
In Evans' book (Page 712), the Gauss-Green theorem is stated without proof and the Divergence theorem is shown as a consequence of it. This may be opposite to what most people are familiar with.