Geodesic equation on a codimension 1 submanifold of $\mathbb{R^{n+1}}$ Section 8.4.3 in Taubes' textbook on Differential Geometry, from which I'm self-studying, is as follows:

What is that convenient coordinate chart that Taubes is referring to? Could someone provide me with a proof that the geodesic equation takes the form that the author describes?
Edit: I should clarify what he means by a geodesic. A geodesic is a curve which satisfies the equation in the following image

 A: I'm not sure if they're exactly the coordinates Taubes is hinting at, but I would choose Cartesian coordinates for $\mathbb R^{n+1}$ centred on a point $p\in \Sigma$ such that $e_1,\ldots,e_n$ is a basis for $T_p \Sigma$. In these coordinates, $\Sigma$ is locally the graph of some function $f : \mathbb R^n \to \mathbb R$ with $f(0)=0, \nabla f(0) = 0;$ so we can use the first $n$ coordinates as a coordinate chart for $\Sigma.$ The plan is to show the two equations are equivalent at the point $p$, using $\nabla f(0) = 0$ to greatly simplify the calculations.
 The metric is  simply $$g_{ij} = (e_i + \partial_i f e_{n+1})\cdot(e_j + \partial_j f e_{n+1}) = \delta_{ij} + \partial_i f\, \partial_j f,$$ and you can check that $$n = \frac{e_{n+1} - \sum_j \partial_j f\, e_j}{\sqrt{1+|\nabla f|^2}}$$ is a unit normal vector field. 
If $\gamma$ is a curve in $\Sigma$ with $\gamma(0)=p$, then its components in $\mathbb R^{n+1}$ are simply $x = (\gamma^1, \ldots, \gamma^n,f(\gamma)).$ Since $\nabla f=0$ at the origin, we see that the derivatives of $g$ vanish there; so $\Gamma(p) = 0$ and thus the geodesic equation at this point is simply $\ddot \gamma(0) = 0.$ On the other hand, differentiating the unit normal and evaluating at $p$ yields $\partial_k n (p) = -\sum_{j} \partial_k \partial_j f\, e_j$; so the equation given in $\mathbb R^{n+1}$ reduces to $$\ddot x(0) = \partial_k \partial_j f(0)\,\dot \gamma^j \dot \gamma^k \, e_{n+1}. \tag 1$$ Differentiating $x = (\gamma(t),f(\gamma(t)))$ twice and evaluating at $p$ we find $$\ddot x(0) = (\ddot \gamma(0), \partial_i \partial_j f(0) \dot \gamma^i \dot \gamma^j);$$ so at $p$, the geodesic equation $\ddot \gamma(0) = 0$ is equivalent to $(1).$
