$\mathbf n$ is the normal of $\Gamma$.
Since $\Delta_S = \nabla_S \cdot \nabla_S$, we deduce
@f[
-\Delta_S v = \Delta \tilde v - \mathbf n^T D \tilde v \mathbf n - (\nabla \tilde v)\cdot \mathbf n (\nabla \cdot \mathbf n).
+\Delta_S v = \Delta \tilde v - \mathbf n^T \ D \tilde v \ \mathbf n - (\nabla \tilde v)\cdot \mathbf n (\nabla \cdot \mathbf n).
@f]
As usual, we are only interested in weak solutions for which we can use $C^0$
solution function. There are (at least) two ways to do that. The first one
is to project away the normal derivative as described above using the natural extension of $u(\mathbf x)$ (still denoted by $u$) over $\mathbb R^d$, i.e. to compute
@f[
- -\Delta_\Gamma u = \Delta u - \mathbf n^T D u \mathbf n - (\nabla u)\cdot \mathbf n (\nabla \cdot \mathbf n).
+ -\Delta_\Gamma u = \Delta u - \mathbf n^T \ D u \ \mathbf n - (\nabla u)\cdot \mathbf n (\nabla \cdot \mathbf n).
@f]
Since we are on the unit circle, $\mathbf n=\mathbf x$ so that
@f[