The free surface is $r=R(x,t)$, which can be rewritten as the zero level set $F(x,r,t) = r-R(x,t) = 0$. Since fluid particles on the free surface must remain on the free surface (this is essentially the no-penetration boundary condition), it follows that the material derivative of $F(x,r,t)$ must be zero, i.e. $$ \frac{DF}{Dt} = \frac{\partial F}{\partial t} + (u,v)\cdot\
abla F = 0 $$ Computing and substituting all the necessary partial derivatives of $F$ into this equation then yields the kinematic condition (2.4). For the (normal) stress boundary condition, it doesn't seem like what I have in mind but perhaps you can figure it out if I explain the terms in equation (2.5)? The vector $(-R_x,1)$ is the (outward) normal vector of the free surface and $R$ in this case corresponds to the radius of curvature. The matrix on the left, if you look close enough, is actually the Newtonian stress tensor.