$\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$Given three linearly independent vectors, $\dot c(t)$, $v(t)$, $\dot v(t)$, and $F(t,s) = c(t) + sv(t)$, we want to calculate $$ \Brak{\frac{\partial^2 F(t,s)}{\partial t\, \partial s},N}, $$ where $N$ is normal vector on $F$.
That is $$ \Brak{\frac{\partial^2 F(t,s)}{\partial t\, \partial s},N} = \frac{ \Brak{\dot v(t), (\dot c(t) + s \dot v(t)) \times v(t)}}{\|(\dot c(t) + s \dot v(t)) \times v(t)\|} = \frac{\Brak{\dot v(t),(\dot c(t))\times v(t)}}{\|(\dot c(t)+s \dot v(t))\times v(t)\|}. $$
My question is, why is the last equality right? That is, why can I write $(\dot c(t) + s\dot v(t)) \times v(t) = \dot c(t) \times v(t)$ when $\dot c(t)$, $v(t)$ and $\dot v(t)$ are linearly independent?
$\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$It's not that $(\dot{c}(t) + s\dot{v}(t)) \times v(t) = \dot{c}(t) \times v(t)$, but that $$ \Brak{\dot{v}(t), s\dot{v}(t) \times v(t)} = 0 $$ because the cross product is orthogonal to each of its "factors". Consequently, $$ \Brak{\dot{v}(t), (\dot{c}(t) + s\dot{v}(t)) \times v(t)} = \Brak{\dot{v}(t), \dot{c}(t) \times v(t)}. $$ This is true whether or not the vectors in question are linearly independent.