So in order to understand Stokes' theorem, I tried to prove it for a parallelogram $P$ with parameterized boundary $R(t)=(x(t),y(t))$ spanned by two vectors $\mathbf{u}$ and $\mathbf{v}$.
The parallelogram :
So that for some covector field "1-form" $\omega$: $$\int_{\partial P}\omega = \int_{P}d\omega$$
With the edges having $t=0,1,2,3$ I think this reduces to: $$\omega_{R(0)}(\mathbf{u})+\omega_{R(1)}(\mathbf{v})-\omega_{R(2)}(\mathbf{u})-\omega_{R(3)}(\mathbf{v})=d\omega(\mathbf{u}\wedge\mathbf{v})$$
And this is my try: $$\mathbf{u}=R(1)-R(0)=R(2)-R(3)\\ \mathbf{v}=R(2)-R(1)=R(3)-R(0)$$
Let $ω_t=ω_{R(t)}$ so: $$\omega_{0}(\mathbf{u})+\omega_{1}(\mathbf{v})-\omega_{2}(\mathbf{u})-\omega_{3}(\mathbf{v})=(ω_0-ω_2)(\mathbf{u})+(ω_1-ω_3)(\mathbf{v})$$
Let: $$ω_1-ω_0=ω_2-ω_3= \Delta_{\mathbf{u}}~ω~,~ω_2-ω_1=ω_3-ω_0= \Delta_{\mathbf{v}}~ω~,\\\Delta_{\mathbf{u}}~ω(\mathbf{v})=\Deltaω(\mathbf{u},\mathbf{v})$$
So: $$(ω_0-ω_2)(\mathbf{u})+(ω_1-ω_3)(\mathbf{v})=\\(-\Delta_{\mathbf{u}} ~ω-\Delta_{\mathbf{v}}~ω)(\mathbf{u})+(\Delta_{\mathbf{u}} ~ω-\Delta_{\mathbf{v}}~ω)(\mathbf{v})=\\ -\Delta_{\mathbf{u}} ~ω(\mathbf{u})-\Delta_{\mathbf{v}}~ω(\mathbf{u})+\Delta_{\mathbf{u}} ~ω(\mathbf{v})-\Delta_{\mathbf{v}}~ω(\mathbf{v})=\\- \Deltaω(\mathbf{u},{\mathbf{u}})-\Deltaω(\mathbf{v},{\mathbf{u}})+\Delta ω(\mathbf{u},{\mathbf{v}})-\Deltaω(\mathbf{v},{\mathbf{v}})$$
I'm not quite sure how to continue from here but if $-\Deltaω(\mathbf{v},{\mathbf{v}})-\Deltaω(\mathbf{u},{\mathbf{u}}) =0$, then let: $$dω(\mathbf{u},\mathbf{v})=\Delta ω(\mathbf{u},{\mathbf{v}})-\Deltaω(\mathbf{v},{\mathbf{u}})$$ It's easy to proof that $d\omega$ is multi-linear and anti-symmetric so: $$d\omega(\mathbf{u},\mathbf{v})=d\omega(\mathbf{u}\wedge \mathbf{v})$$
So I know I probably made a lot of mistakes but is the core of this way to prove it right? Are there any papers who proof Stokes' Theorem in general by proving it for a parallelogram or parallelepiped spanned by vectors? Because it makes much sense and it's a beautiful approach, but what are my exact mistakes?