I am trying to compute the pushforward of a tangent vector as a derivation. Here are my definitions:
Let $v\in \left.(TM)\right|_U$ be a tangent vector at the point $p$. Given some coordinates $(\phi,U)$, this can be expressed as \begin{align} v=\sum_{i}a^i\frac{\partial}{\partial \phi^i}(p). \end{align}
Define the pushforward of a vector under a map $\psi$ by its action on a function $f$: \begin{align} (\psi_*v)(f)=v(f\circ \psi). \end{align}
What I want to calculate is the pushforward of $v$ under the chart $\phi$. As far as I can understand from reading, this should give me the vector $a$, but I'm struggling to get there. I think it is something to do with exactly how the standard unit vectors in $\mathbb{R}^n$ are defined, as I'm not sure. Here is my attempt:
\begin{align} (\phi_*v)(f)&=v(f\circ \phi)\\ &=\sum_{i}a^i\frac{\partial}{\partial \phi^i}(p)(f\circ \phi)\\ &=\sum_{i}a^i\frac{\partial (f\circ \phi)}{\partial \phi^i}(p)\\ &=\sum_{i}a^i D_i(f\circ \phi\circ \phi^{-1})(\phi(p))\\ &=\sum_{i}a^i D_i(f)(\phi(p)). \end{align}
The third to fourth line is the defintion of the 'partial derivative' notation.
$f$ is ultimately a test function, so I really want to take it out. The last line makes me think I want something like \begin{align} \sum_{i}a^i D_i(f)(\phi(p))=\left(\sum_{i}a^i D_i(\phi(p))\right)f, \end{align} in which case I can see if I think of $D_i(\phi(p))$ as the standard unit basis of $\mathbb{R}^n$, then this gives me the vector $a$ as I want.
Is this correct? Is $D_i(\phi(p))$ standard notation? Any help would be appreciated.