A space vector $\mathbf{f}$ can be expressed in orthogonal curvilinear component form as,
$\mathbf{f} = \mathbf{i}_1 \, f_1 + \mathbf{i}_2 \, f_2 + \mathbf{i}_3 \, f_3 = \mathbf{f}_n + \mathbf{f}_s$,
where: $\mathbf{f}_n \equiv \mathbf{n}\mathbf{n} \cdot \mathbf{f} = \mathbf{i}_3 \, f_3$ is its normal part, and $\mathbf{f}_s \equiv \mathbf{I}_s \cdot \mathbf{f} = \mathbf{i}_1 \, f_1 + \mathbf{i}_2 \, f_2$ the tangential to the surface whose unit tangent vectors are $(\mathbf{i}_1, \mathbf{i}_2)$, and being $\mathbf{I}_s = \mathbf{I} - \mathbf{n}\mathbf{n}$, is the dyadic surface idemfactor, and $\mathbf{I}$ the three-dimensional spatial idemfactor.
It is correct $(\nabla_s \mathbf{f}) \cdot \mathbf{I}_s = \nabla_s \mathbf{f}_s$?