I'm reading through the proof in this paper (http://www.math.leidenuniv.nl/scripties/JaibiBach.pdf) but I'm stuck at the line:
"Using the chain rule we get:
$L_p(\phi_v) = -Dn(\phi_v) = - \frac {\partial}{\partial v} (n \circ \phi)$"
How does the second equality follow by the chain rule?
My working is: $-Dn(\phi_v) = - \frac{d}{dt} (n \circ \frac {\partial \phi}{\partial v}) = \frac {dn}{dt} ((\frac {\partial \phi}{\partial v})(p)) \frac {d} {dt}( \frac {\partial \phi}{\partial v}) (p) $ but I don't know how to continue this.
No, you're misunderstanding the notation. The author is evaluating the derivative of $n$ at $p$ (that was omitted) on the tangent vector $\phi_v$. The chain rule tells us that $$Df_p(\xi) = \frac d{dt}\Big|_{t=0} f(g(t)) \quad \text{for any curve } g \text{ in $M$ with } g(0)=p \text{ and } g'(0)=\xi\,.$$ In this case, he's just taking the path $g(t) = \phi(u_0,t)$ and we get the partial derivative with respect to $v$.