Second derivative of vector flows: if $g(t)=\varphi_{v_1}^t\circ \varphi_{v_2}^t-\varphi_{v_2}^t\circ \varphi_{v_1}^t$ then $g''(0)=-2[v_1,v_2](p)$

Question

Let $v_1$ and $v_2$ be two vector fields on a manifold $M$, and let $g(t)=\varphi_{v_1}^t\circ \varphi_{v_2}^t-\varphi_{v_2}^t\circ \varphi_{v_1}^t$. I want to prove that $$g''(0)=-2[v_1,v_2](p).$$ To do this, I tried writing $$v_1(p)=\sum_{j=1}^n a_j(p)\frac{\partial}{\partial x_j}\Big\vert_p \quad \text{and} \quad v_2(p)=\sum_{j=1}^n b_j(p)\frac{\partial}{\partial x_j}$$ Using that $\frac{\partial}{\partial t}\varphi^t_{v_i}=v_i(\varphi^t_{v_i})$ for $i=1,2$ I get that the first derivative is $$g'(t)=v_1(\varphi_{v_1}^t(\varphi_{v_2}^t(p)))\cdot v_2(\varphi_{v_2}^t(p)))-v_2(\varphi_{v_2}^t(\varphi_{v_1}^t(p)))\cdot v_1(\varphi_{v_1}^t(p)))$$ This is my attempt for the second derivative: $$g''(t)= \frac{\partial}{\partial p}v_1(p)\Big|_{\varphi_{v_1}^t(\varphi_{v_2}^t(p))}\cdot v_2(\varphi_{v_2}^t(p)))+v_1(\varphi_{v_1}^t(\varphi_{v_2}^t(p)))\cdot \frac{\partial}{\partial p}v_2(p)\Big|_{\varphi_{v_2}^t(p))}- \frac{\partial}{\partial p}v_2(p)\Big|_{\varphi_{v_2}^t(\varphi_{v_1}^t(p))}\cdot v_1(\varphi_{v_1}^t(p)))-v_2(\varphi_{v_2}^t(\varphi_{v_1}^t(p)))\cdot \frac{\partial}{\partial p}v_1(p)\Big|_{\varphi_{v_1}^t(p))}$$ However, as $\varphi^0_{v_i}(p)=p$ for $i=1,2$, I get that $g''(0)=0$. What am I doing wrong?