= Solution
In <Einstein summation convention>, the <curl> of a <cross product> has components
$$
\begin{aligned}
[\nabla\times(a\times b)]_i
&=\epsilon_{ijk}\partial_j(\epsilon_{k\ell m}a_\ell b_m)\\
&=\partial_j(a_i b_j-a_j b_i)\\
&=a_i\partial_jb_j-b_i\partial_ja_j+b_j\partial_ja_i-a_j\partial_jb_i.
\end{aligned}
$$
The second line uses the <contraction of two Levi-Civita symbols>; the last uses the <product rule>. These components prove the <curl of a cross product> identity, including both <divergence> and directional-derivative terms.
Now apply that identity to $m\times v$. On the <orientable surface>, $m=n$, $m\cdot v=0$, and $|m|=1$. Differentiating $m\cdot m=1$ gives $m_i\partial_jm_i=0$. Taking a <dot product> with $m$ in the <curl of a cross product> formula therefore gives, on $S$,
$$
\begin{aligned}
n\cdot\nabla\times(m\times v)
&=\nabla\cdot v-(m\cdot v)\nabla\cdot m
+m_i v_j\partial_jm_i-m_i m_j\partial_jv_i\\
&=(\delta_{ij}-n_in_j)\partial_jv_i.
\end{aligned}
$$
Only tangency on the surface is needed; no claim is made that the extension $v$ is tangent away from it. The right side is the <surface divergence> of the tangent <vector field>.
With the boundary oriented according to <Stokes theorem>, cyclic invariance of the <scalar triple product> gives
$$
(m\times v)\cdot d\mathbf r=v\cdot(d\mathbf r\times n)=v\cdot u\,ds,
$$
where $u$ is the <outward conormal of a surface boundary>. Thus <Stokes theorem> proves
$$
\boxed{\int_S(\delta_{ij}-n_in_j)\partial_jv_i\,dS
=\oint_C u\cdot v\,ds.}
$$
This is the <surface divergence theorem for a tangent field>.
For the disc in the $z=0$ plane, take $n=(0,0,1)$ and $v=\mathbf r=(x,y,z)$. The field is tangent on the disc, and $\partial_jv_i=\delta_{ij}$. The contraction is $3-|n|^2=2$, so the <surface integral> is $2\pi R^2$. On the positively oriented circular boundary, $u=(\cos\theta,\sin\theta,0)$ and $v=Ru$, so the boundary <flux> is $R(2\pi R)=2\pi R^2$. \b[Both sides agree.]
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