Let $\Omega \subset \mathbb{R}^n$ be an open, bounded domain with Lipschitz boundary and $A=\{a_{ij}\} \subset C^{0,1}(\Omega)$ be a symmetric matrix with Lipschitz continuous entries that is also uniformly elliptic:
$$A(x) \xi \cdot \xi \geq \lambda |\xi|^2, \qquad \forall x \in \Omega, \quad \forall \xi \in \mathbb{R}^n$$
for some $\lambda>0$. Let also $c \in L^\infty(\Omega)$. Assume $u \in H^2(\Omega)$ solves the problem
$$ \begin{cases} div(A \nabla u) + cu = 0, \qquad &\text{in} \ \Omega \\ u = 0, & \text{on} \ \partial \Omega \\ A \nabla u \cdot \nu = 0, & \text{on} \ \partial \Omega \end{cases} $$
where $\nu$ is the unit outward normal. I want to conclude that $u\equiv 0$ in $\Omega$. Upon searching the internet it turned out that such questions are referred to as unique continuation principle. Specifically I came across these notes, where Theorem 1.4 implies that if $u$ solves the PDE and $u = \partial_\nu u = 0$ on $\partial \Omega$, then $u \equiv 0$. In above case instead of $\partial_\nu u$ we have $A \nabla u \cdot \nu$. I was wondering if the result should still be true. Also why does one consider $\partial_\nu u$ for the unique continuation principle instead of $A \nabla u \cdot \nu$? I thought the latter was more natural as it appears in the weak formulation of the problem.
For simplicity let's take $u \in C^2(\Omega) \cap C^1(\overline{\Omega})$. Since $u=0$ on $\partial \Omega$, $\nabla u$ is normal to $\partial \Omega$ so $\nabla u = (\partial_\nu u) \nu$. Hence, $$0=A\nabla u \cdot \nu = (\partial_\nu u) A\nu \cdot \nu .$$ By ellipticity, $A\nu \cdot \nu \geqslant \lambda >0$, so $$ \partial_\nu u =0 \qquad \text{ on } \partial \Omega.$$ Hence, you're in the exact same scenario as in your lecture notes, so $u \equiv 0$. This remains to be proven for $u \in H^2(\Omega)$ but I believe the above computation shows that the result will be true.
As for your question about $\partial_\nu u =0$ vs $A\nabla u \cdot \nu =0$ on $\partial \Omega$, sure $A\nabla u \cdot \nu =0$ appears in the weak formulation but boundary conditions are motivated by the physical application. The condition $\partial_\nu u =0$ has a physical meaning (no flux through the boundary) where as $A\nabla u \cdot \nu =0$ has no clear physical interpretation.