Question: Let $\gamma \colon [0,1] \to \mathbb{R}^2$ be so that it is continuous and piecewise differentiable and defines a simple closed graph on $\mathbb{R}^2$ that encloses a bounded open connected set $\Omega$. Let $u$ be continuously differentiable on an open connected set that contains the closure of $\Omega$. Show that there exists a function on the path of $\gamma$, defined as $v \colon \gamma([0,1]) \to \mathbb{R}^2$ so that $$\int_\Omega \frac{\partial u}{\partial x} d A = \int_{\partial \Omega} u(\sigma) \cdot v (\sigma) \, d \sigma.$$ Hence, show that there exists $c_0 \in \mathbb{R}$ with $|c_0| = 1$ so that $$\int_\Omega \frac{\partial u}{\partial x} d A = c_0 \int_0^1 u(\gamma(t)) \gamma_2'(t) \, dt$$ where $\gamma(t) = (\gamma_1(t), \gamma_2(t))$.
My attempt: For the first part, it seems like a direct application of divergence theorem, but I don't understand why should the normal vector be defined on the path of $\gamma$. As for the second part, I am unsure how to make the change of variable. Any hints are appreciated.
Gauss' divergence theorem says \begin{align} \int_\Omega{\rm div}\,\mathbf{u}\,dA&=\int_\Omega\partial_x u_1+\underbrace{\partial_yu_2}_{(*)}\,dA=\oint_{\partial\Omega}\mathbf{u}\cdot\mathbf{n}\,d\sigma\\ &=\int_0^1-\underbrace{u_1(\gamma(t))\,\dot\gamma_2(t)}_{(**)}+u_2(\gamma(t))\,\dot\gamma_1(t)\,dt\,.\tag{1} \end{align} where we use the fact that the normal vector $\mathbf{n}$ at $\partial \Omega$ is perpendicular to the tangent $\dot\gamma\,,$ hence parallel to $$ \begin{pmatrix}-\dot\gamma_2\\\dot\gamma_1\end{pmatrix}\,. $$ Green's theorem says $$\tag{2} \int_\Omega\partial_xu_2-\partial_yu_1\,dA=\oint_{\partial\Omega}\mathbf{u}\cdot\,d\mathbf{s}= \int_0^1u_1(\gamma(t))\,\dot\gamma_1(t)+u_2(\gamma(t))\,\dot\gamma_2(t)\,dt\,. $$ If we apply Green's theorem to $(u_2,u_1)$ instead of $\mathbf{u}$ we get \begin{align}\tag{3} \int_\Omega \partial_xu_1-\underbrace{\partial_yu_2}_{(*)}\,dA=\int_0^1u_2(\gamma(t))\,\dot\gamma_1(t)+\underbrace{u_1(\gamma(t))\,\dot\gamma_2(t)}_{(**)}\,dt\,. \end{align} Adding (1) and (3) and cancelling the underbraced terms with opposite signs gives $$\tag{4} \int_\Omega\partial_xu_1\,dA=\int_0^1u_2(\gamma(t))\,\dot\gamma_1(t)\,dt=\int_{\partial \Omega}\mathbf{u}\cdot\mathbf{v}\,d\sigma $$ where $$ \mathbf{v}=\begin{pmatrix}0\\\dot\gamma_1\end{pmatrix} $$ is the projection of the normal $\mathbf{n}$ at $\partial \Omega$ onto the $y$-axis. This solves the first task.