$\mathbf n$ 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 - \mathbf n \ D \mathbf n \ \mathbf n ).
+\Delta_S v = \Delta \tilde v - \mathbf n^T \ D^2 \tilde v \ \mathbf n - (\mathbf n \cdot \nabla \tilde v) (\nabla \cdot \mathbf n - \mathbf n^T \ D \mathbf n \ \mathbf n ).
@f]
Worth mentioning, the term $\nabla \cdot \mathbf n - \mathbf n \ D \mathbf n \ \mathbf n$ appearing in the above expression is the total curvature of the surface (sum of principal curvatures).
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 \kappa,
+ -\Delta_\Gamma u = \Delta u - \mathbf n^T \ D^2 u \ \mathbf n - (\mathbf n \cdot \nabla u)\ \kappa,
@f]
where $\kappa$ is the total curvature of $\Gamma$.
Since we are on the unit circle, $\mathbf n=\mathbf x$ and $\kappa = 1$ so that