I am relatively new to advanced math so sorry in advance if the following is not phrased correctly or not rigorously enough.
Let $M_1, M_2$ be differentiable manifolds, $\phi:M_1\rightarrow M_2$ be a smooth map and the mapping $f:M_2 \rightarrow \mathbb R$. Also, let $p \in M_1$ and $v \in T_pM_1$. Now, consider the push-forward $\phi_*: T_pM_1 \rightarrow T_{\phi(p)}M_2$ defined as $\phi_*(v)=d\phi_p(v)$ with $d\phi$ being the differental of $\phi$.
Lastly, consider the pull-back $\phi^*(f)=f \circ \phi$, or more clearly $(\phi^*(f))(p)=f(\phi(p))$.
Now, the question is:
Upon giving us the above definition for the pushover, which just sends tangent vectors at $p$ in $T_pM_1$ to tangent vectors at $\phi (p)$ in $T_{\phi (p)}M_2$, our professor gave us another definition for the push-forwards which relates it to the pullback. The definition is as follows:
$(\phi_*(v))(f)(\phi (p))=v(\phi^*(f))(p)=(u(f \circ \phi))(p)$.
So, finally, my problem is that I can't quite get the geometric intuition behind their relation. I try to view it in terms on mappings, and while I get why the pullback is called the pullback(check the image in the 2nd answer in this link: Geometric intuition behind pullback?), and I get the intuition behind what the push-forward does(which I described above), I don't know how to relate them.
I mean, when we use the push-forward, we are pushing tangent vectors in $M_1$ over to tangent vectors in $M_2$. But, why is the last definition that I gave valid(well, it's a definition, so it might be better to ask why is it intuitive)?
In particular, I would like an intuitive explanations using mapping from and to $M_1, M_2, \mathbb R$.
EDIT: The above notation follows closely the notation from Do Carmo's "Riemannian Geometry".