I want to show given $v\in T_pM$, then $D_vX$ is completely determined by $X$ restricted to any curve $r$ with $r'(0)=v,r(0)=p$.
I have shown that if $r_1'(0)=r'_2(0)=v,$ then $D_{r_1'(0)}X=D_{r_2'(0)}X$ by expressing $X,v$ using local coordinate $(\frac{\partial}{\partial x^i})$ around $p$ and apply the axiom of a connection.
But I have trouble showing the other part, that is, given $r:[0,\epsilon)\to M$ be a curve with $r'(0)=v$, and $X,Y$ be two vector fields such that $X(r(t))=Y(r(t))\forall t\in[0,\epsilon)$, then $D_vX=D_vY$.
Since by definition, $D_vX=D_VX(p)$, I don't know how and why we can restrict $X$ on $r(t)$.
This refers to the definition of $\bigtriangledown_{r'}r'$, which is used in geodesic.
Proof We consider the local presentation of $\bigtriangledown$ $$V=V_i\frac{\partial}{\partial v_i},~~X=X_i\frac{\partial}{\partial v_i},~~Y=Y_i\frac{\partial}{\partial v_i},~~r'=\frac{dv_i}{dt}\frac{\partial}{\partial v_i}$$
Because $V_p=r'(0)$, we have $$V_i(p)=\frac{dv_i}{dt}|_0~~~\cdots(1)$$
From $X\circ r(t)=Y\circ r(t)$, we get $$X_i\circ r(t)=Y_i\circ r(t)~~~\cdots(2)$$
Then we have $$\bigtriangledown_VX(p)=(V_i(p)X_j(p)\Gamma_{ij}^k(p)+V_i(p)\frac{\partial X_k}{\partial v_i}|_p)\frac{\partial}{\partial v_k}|_p$$ $$=(V_i(p)Y_j(p)\Gamma_{ij}^k(p)+\frac{dv_i}{dt}|_0\frac{\partial X_k}{\partial v_i}|_p)\frac{\partial}{\partial v_k}|_p~~~\cdots\text{From (1)}$$ $$=(V_i(p)Y_j(p)\Gamma_{ij}^k(p)+\frac{dX_k\circ r}{dt}|_0)\frac{\partial}{\partial v_k}|_p~~~\cdots\text{From chain rule}$$ $$=(V_i(p)Y_j(p)\Gamma_{ij}^k(p)+\frac{dY_k\circ r}{dt}|_0)\frac{\partial}{\partial v_k}|_p~~~\cdots\text{From (2)}$$ $$=(V_i(p)Y_j(p)\Gamma_{ij}^k(p)+\frac{dv_i}{dt}|_0\frac{\partial Y_k}{\partial v_i}|_p)\frac{\partial}{\partial v_k}|_p$$ $$=(V_i(p)X_j(p)\Gamma_{ij}^k(p)+V_i(p)\frac{\partial Y_k}{\partial v_i}|_p)\frac{\partial}{\partial v_k}|_p$$ $$=\bigtriangledown_VY(p)$$