What is the physical meaning of $v\times n$? What is the physical meaning of $v\times n$, where $v$ is a velocity vector and and $n$ is a unit normal vector of a interface? Why at the free surface between two fluids,
$$v^{(1)}\times n=v^{(2)}\times n,$$
where $v^{(1)}$ and $v^{(2)}$ are velocity vectors on both sides of the surface. 
Problem solved.
 A: This condition means that the tangential components of the velocities of the two fluids are the same at their interface. It is equivalent to saying that there is no slip between the two fluids.
Equality of tangential components is not restricted to fluid dynamics alone. You can find them in electrodynamics as well, where the tangential component of the $\vec{E}$ field are required to be continuous across an interface while that of $\vec{H}$ are discontinuous in the presence of surface currents.
A: It is a common misunderstanding to say that $\hat{n} \times \mathbf{v}$ is the tangential or transverse velocity.  This vector is orthogonal to both $\mathbf{v}$ and $\hat{n}$, so it is not a component of either vector.  The transverse-to-$\hat{n}$ component of $\mathbf{v}$ is defined as:
$$
\mathbf{v}_{t} = \hat{n} \times \left( \mathbf{v} \times \hat{n} \right)
$$
The form you show is derived from an assumption of boundary conditions.  In this particular case, it looks like a conservation of vorticity across a boundary, where $\nabla \times \mathbf{v}$ $\rightarrow$ $\hat{n} \times \mathbf{v}$.
Another form of this type is used for plane electromagnetic waves.  One assumes that $\delta \mathbf{B}$ and $\delta \mathbf{E}$ are orthogonal to the propagation direction $\mathbf{k}$.  Then using Maxwell's equations and some assumptions, one can then show that:
$$
\nabla \cdot \delta \mathbf{B} \rightarrow \hat{n} \cdot \left( \delta \mathbf{B}_{2} - \delta \mathbf{B}_{1} \right) = 0 \\
\nabla \times \delta \mathbf{E} \rightarrow \hat{n} \times \left( \delta \mathbf{E}_{2} - \delta \mathbf{E}_{1} \right) = 0
$$
The first equation derives from the assumption that there are no magnetic monopoles (I think there are more subtleties involved too, but this is easy enough for now) and the second from the first, since $\nabla \times \delta \mathbf{E}$ = $- \partial_{t} \mathbf{B}$ and that one can show $\oint \mathbf{E} \cdot d\mathbf{l} = 0$ here.
