Let $\gamma$ be an arclength parametrization of the curve, so that $\Delta_{\rm LB} u = \frac{d^2}{dt^2}(u \circ \gamma)$. From the definition of the curve we have $s \circ \gamma = 0$, so differentiating this twice we get the equations
$$\langle \nabla s , \dot \gamma \rangle = 0 \; \text{and} \;
\nabla^2 s(\dot \gamma, \dot \gamma) + \langle \nabla s, \ddot \gamma\rangle = 0.
$$
Since $\gamma$ is unit-speed we know $\langle \dot \gamma, \ddot \gamma \rangle = 0$ and thus $\ddot \gamma$ and $\nabla s$ are parallel; so putting these together we get
$$\ddot \gamma = \frac{\langle\ddot \gamma, \nabla s\rangle \nabla s }{|\nabla s|^2 }
= \frac{-\nabla^2 s(\dot\gamma,\dot\gamma)}{|\nabla s|^2}\nabla s.
$$
Since $|\dot\gamma|=1$ we can assume without loss of generality that $\dot \gamma = \frac{J \nabla s}{|\nabla s|}$, where $J$ is either a clockwise or counterclockwise rotation by $\pi/2$. (It turns out not to matter which.) We can now write $\dot \gamma$ and $\ddot \gamma$ in terms of the derivatives of $s$, so the Laplacian is
$$ \frac{d}{dt}( \nabla u \cdot \dot \gamma ) = \nabla^2 u( \dot \gamma, \dot \gamma) + \nabla u \cdot \ddot \gamma = \frac{\nabla^2 u ( J \nabla s, J \nabla s ) }{|\nabla s|^2} - \frac{\nabla^2 s(J \nabla s,J \nabla s)}{|\nabla s|^4}\langle\nabla s,\nabla u\rangle.$$
If we define the operator $Q(f) = \frac{\nabla^2 f ( J \nabla s, J \nabla s ) }{|\nabla s|^2}$ (which is simply the second derivative of $f$ in the direction tangent to $\gamma$) then this is easier to understand:
$$ \Delta_{\rm LB} u = Q(u) - Q(s)\frac{\langle \nabla s, \nabla u \rangle}{|\nabla s|^2}.$$
The first term is the second derivative along the line tangent to the level curve of $s$ and the second term accounts for the deviation of the level curve from this straight line.
