Let $\mathbf{A}$ be a vector function in $\mathbf{R}^3$ and we want to find the normal and tangent components of $\nabla \times \mathbf{A}$ on a smooth and closed surface $\Gamma$. $\mathbf{n}$ is the unit vector normal to the surface $\Gamma$.
In order to find the tangent and normal components of, we need to define the surface operators $\nabla_t$ and $\nabla_n$. The normal gradient is $\nabla_n = \mathbf{n(n} \cdot \nabla)$. The surface gradient is $\nabla_t = \nabla - \mathbf{n(n} \cdot \nabla)$.
Now, let's expand $\nabla \times \mathbf{A} = (\nabla_t + \nabla_n) \times (\mathbf{A}_t + \mathbf{A}_n) = \nabla_t \times \mathbf{A}_t + \nabla_t \times \mathbf{A}_n + \nabla_n \times \mathbf{A}_t + \nabla_n \times \mathbf{A}_n$
Now, let's expand each of the terms. We can use the regular identities to expand with surface operators.
$\nabla_t \times \mathbf{A}_t$ is expanded using the identity $\nabla \cdot (\mathbf{A \times B) = (\nabla \times A)\cdot B - (\nabla \times B)\cdot A}$. Let $\mathbf{A = n}$ and $\mathbf{B} = \mathbf{A}_t$. Thus, we obtain the following (apply the property $\nabla_t \times \mathbf{n} = 0$, which is true for any surface):
$\nabla_t \times \mathbf{A}_t = -\nabla_t \cdot ( \mathbf{n} \times \mathbf{A}_t ) \mathbf{n}$
$\nabla_t \times \mathbf{A}_n$ is expanded using the identity $\nabla \times (k\mathbf{A}) = k(\nabla \times \mathbf{A}) + \nabla k\times \mathbf{A}$
$\nabla_t \times \mathbf{A}_n = \nabla_t \times [(\mathbf{A}_n \cdot \mathbf{n}) \mathbf{n}] = (\mathbf{A}_n \cdot \mathbf{n})(\nabla_t \times \mathbf{n}) + [\nabla_t (\mathbf{A}_n \cdot \mathbf{n}) ]\times \mathbf{n} = [\nabla_t (\mathbf{A}_n \cdot \mathbf{n}) ]\times \mathbf{n}$
$\nabla_n \times \mathbf{A}_t$ requires the use of the definition of $\nabla_n = \mathbf{n}(\mathbf{n}\cdot \nabla) = \mathbf{n}(\frac{\partial}{\partial n})$, as follows:
$ \nabla_n \times \mathbf{A}_t = \Big ( \mathbf{n} \frac{\partial}{\partial n} \Big) \times \mathbf{A}_t = \mathbf{n} \times \Big ( \frac{\partial}{\partial n} \mathbf{A}_t \Big) $
The term $\nabla_n \times \mathbf{A}_n$ is expanded as follows:
$ \nabla_n \times \mathbf{A}_n = \Big ( \mathbf{n} \frac{\partial}{\partial n} \Big) \times \mathbf{A}_n = \mathbf{n} \times \Big ( \frac{\partial}{\partial n} \mathbf{A}_n \Big ) = \mathbf{n} \times \Big [ \Big ( \frac{\partial}{\partial n} A_n \Big ) \mathbf{n} + A_n \frac{\partial}{\partial n} \mathbf{n} \Big ] $ $= \mathbf{n} \times \Big [ \Big ( \frac{\partial}{\partial n} A_n \Big ) \mathbf{n} + A_n 0 \Big ] = 0 $.
In the last expression, the identity $\frac{\partial}{\partial n} \mathbf{n} = 0$ was used. The proof for the latter is as follows. First, consider the identity $\nabla (\mathbf{A \cdot B}) = (\mathbf{A} \cdot \nabla) \mathbf{B} + (\mathbf{B} \cdot \nabla ) \mathbf{A} + \mathbf{A} \times( \nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})$. Now, let $\mathbf{A = B = n}$ to get the following:
$ \nabla (\mathbf{n \cdot n}) = (\mathbf{n} \cdot \nabla) \mathbf{n} + (\mathbf{n} \cdot \nabla ) \mathbf{n} + \mathbf{n} \times( \nabla \times \mathbf{n}) + \mathbf{n} \times (\nabla \times \mathbf{n}) $ $ \rightarrow 0 = 2\frac{\partial}{\partial n}\mathbf{n} + \mathbf{n} \times 0 + \mathbf{n} \times 0 $ $ \rightarrow \frac{\partial}{\partial n}\mathbf{n} = 0 $
Finally, we get the following for $\nabla \times \mathbf{A}$:
$\nabla \times \mathbf{A} = \nabla_t \times \mathbf{A}_t + \nabla_t \times \mathbf{A}_n + \nabla_n \times \mathbf{A}_t + \nabla_n \times \mathbf{A}_n $ $ = -[\nabla_t \cdot (\mathbf{n} \times \mathbf{A}_t )] \mathbf{n} + [\nabla_t (\mathbf{A}_n \cdot \mathbf{n}) ]\times \mathbf{n} + \mathbf{n} \times \Big ( \frac{\partial}{\partial n} \mathbf{A}_t \Big)$.
Finally, the normal component of $\nabla \times \mathbf{A}$ is simply $-[\nabla_t \cdot (\mathbf{n} \times \mathbf{A}_t )] \mathbf{n}$.
The divergence theorem for surfaces is as follows (see page 1029 in Electromagnetic Fields of Jean Van Bladel):
$\int_s \nabla_t \cdot \mathbf{P} dS = \int_c \mathbf{P}\cdot \mathbf{m} dc - \int_s (\nabla_t \cdot \mathbf{n}) (\mathbf{P}\cdot \mathbf{n})dS$, where $\mathbf{m}$ is a vector perpendicular to a contour $c$ on the surface $\Gamma$. This integral vanishes to zero if $\mathbf{P}$ is tangent to $\Gamma$. Let's find the following integral:
$\int (\nabla \times \mathbf{A})_n \cdot \mathbf{n} dS = \int -[\nabla_t \cdot (\mathbf{n} \times \mathbf{A}_t )] dS$, which is zero according to the divergence theorem for surfaces.
Notice that the vector $\mathbf{A}$ and the surface $\Gamma$ are arbitrary at this point. I know that the previous result with the divergence theorem does not mean that locally $(\nabla \times \mathbf{A})_n = 0$, but it certainly means that the integral along the surface of $(\nabla \times \mathbf{A})_n$ is zero for any vector in any surface.
I wonder if this result is correct? I am trying to make analogies with a magnetic field defined through a magnetic vector potential $\mathbf{B} = \nabla \times \mathbf{A}$, but I don't see why $\int_S \mathbf{B}\cdot \mathbf{n} dS$ would be zero for any $\mathbf{A}$ in any surface.