Let ($U, F, V$) be a local parametrization of a regular surface $S$ at $p \in S$ with $F(u_0) = p$, $u_0 \in U$.
(1) Find a local parametrization $(U_1, F_1, V_1)$ of $S$ at $p$ such that $F_1(0, 0) = p$.
(2) Let $(x_1, x_2)$ be an orthonormal basis of $T_pS$. Find a local parametrization $(U_2, F_2, V_2)$ of $S$ at $p$ with $F_2(0,0) = p$ and such that the matrix of the first fundamental form with respect to $(U_2, F_2, V_2)$ is the identity matrix.
How do I approach this problem?
How do we calculate local parametrizations?
What does the orthonormal basis is part (2) tell us when finding a local parametrization?
$I$: Here $F: U \subset \mathbb{R}^2 \to F(U):=V \subset S$ where $u_0 \in U$ and $F(u_0) = p \in S$. You just want a coordinate system in which $u_0$ is the origin. Let $(u',v') = (u_1-u,v_1-u)$ where we define $u_0= (u_1,v_1)$. Hence we have a map $\Phi(u,v) = (u',v')$ and now take $F_1= F \circ \Phi$ then,
$$F_1(0,0) = F(u_0) = p$$
i.e you have the triple $(U_1=U \cap \Phi(U'), F_1, (F \circ \Phi)(U'):=V_1)$ where $U' = \{(u',v')\}$.
$II$: The first fundamental form is given by $|\sigma_u|^2 du^2 +2 \sigma_u \cdot \sigma_v\ dudv + |\sigma_v|^2 dv^2$. You want $|\sigma_u|^2= |\sigma_v|^2=1$ and $\sigma_u \cdot \sigma_v = 0$. Since you have the orthornormal basis $x_1,x_2$ for $T_pS$, then they are velocity vectors of curves $\gamma_i: (- \epsilon_i, \epsilon_i) \to S$ where $i = 1,2$ with $\gamma_i(0) = p$.
$\bullet$ Show that there exists a chart $(F,U) = (F,(u,v))$ with $F_u = x_1$ and $F_v = x_2$. Here the existence of the orthornormal basis says $x_1,x_2$ are unit vectors and $x_1 \cdot x_2 = 0$ which is precisely what you need.