Let $E, F \to M$ be two vector bundles based at $M$ and $\pi: T^*M \to M$ the cotangent bundle. For any $\ell \in \mathbb Z$, we set $$\text{Smb}_\ell (E, F) = \{\sigma \in \text{Hom}(\pi^*E, \pi^*F)~|~\sigma(m, \lambda \xi) = \lambda^\ell \sigma(m, \xi),~\forall (m, \xi) \in T^*M\},$$ where $\pi^*E$ is a bundle based at $T^*M$. We call symbol of a linear differential operator $L: \Gamma(E) \to \Gamma(F)$ of order at most $\ell$ the section $\sigma(L) \in \Gamma(\text{Smb}_\ell (E, F))$ such that $$L = \sum_{|\alpha| \le \ell} a_\alpha \partial^\alpha \quad \Rightarrow \quad \sigma(L)(m, \xi) := i^\ell \sum_{|\alpha| = \ell} a_\alpha(m) \xi^\alpha.$$ Now, in my Global Analysis course, we have seen another way to define the symbol. Consider $(m, \xi) \in T^*M$ and $p \in E_m$, pick $f \in C^\infty(M)$ and $s \in \Gamma(E)$ such that $f(m) = 0$, $df_m = \xi$ and $s(m) = p$. We define $$\sigma(L)(m, \xi)p = L\left(\frac{i^\ell}{\ell!}f^\ell s\right)\bigg|_m \in F_m.$$ I tried to do the computation to get back to the previous definition of symbol we had but I do not get the right answer.
Here is what I've done: We have \begin{align} L\left(\frac{i^\ell}{\ell!}f^\ell s\right) &= \frac{i^\ell}{\ell!} \sum_{|\alpha| \le \ell} a_\alpha \partial^\alpha(f^\ell s)\\ &= \frac{i^\ell}{\ell!} \sum_{|\alpha| \le \ell} a_\alpha \sum_{|\beta| \le |\alpha|}\underbrace{\partial^\beta(f^\ell)}_{A_\ell^{\ell - |\beta|}f^{\ell - |\beta|} \partial^\beta f} \partial^{\alpha - \beta}s \end{align} with $A_\ell^{\ell - |\beta|} = \ell(\ell - 1) \ldots (\ell - |\beta|)$. Now, by evaluating this expression at $m$ and using the fact that $f(m) = 0$, the only term that does not vanish is $|\beta| = |\alpha| = \ell$ so that \begin{align} L\left(\frac{i^\ell}{\ell!}f^\ell s\right)\bigg|_m &= i^\ell \sum_{|\alpha| = \ell}a_\alpha(m)(p) \partial^\alpha f. \end{align} Now I feel that I am not far from the answer but I am unable to conclude. Could one of you help me to finish this argument ?