Why we can without loss of generality assume that $\nabla_{Y_p}X=0$ for all $Y_p\in T_pM$?

Question

I know that $\nabla_{Y}X|_p=\nabla_{Y_p}X_{u_p}$ where $u_p$ is a neighborhood around $p$.

Q1: Suppose $p$ is a critical point of $f:M\to \Bbb R$ then why $X_pXf$ depends only on $X_p$? Is that true $X_pYf$ for arbitrary $Y$?

By above conditions

Q2: Why we can without loss of generality assume that $\nabla_{Y_p}X=0$ for all $Y_p\in T_pM$?

Answer 1

If your vector fields $Y$ and $Z$ are such that $Y_p=Z_p=V$, then since $p$ is a critical point of $f$, you have

$$V(Zf)-V(Yf)=Y_p(Zf)-Z_p(Yf)=[Y,Z]_pf=df_p([Y,Z]_p)=0.$$

Thus, $X_p(Xf)$ only depends of the value of $X_p$.