Let $M_1, M_2$ two smooth manifolds, $F: M_1 \rightarrow M_2$ a smooth map and $f:M_2 \rightarrow \mathbb{R}$ a smooth function. I'm trying to prove that for any vector field $V$ on $M_1$, we have $V(F^*f)=DF(V)(f)$.
I know that $f$ is a 0-form and so we have $F^*f=f \circ F$, by definition. But then I don't really know what to do with that.
Moreover, I'm a bit confused with some point related to the application of $V$ on this pullback. As far as I know, $V$ sends a point $x \in M_1$ to some vector $V_x \in T_xM_1$. But then, since $f \circ F$ is a map from $M_1$ to $\mathbb{R}$, we are not applying $V$ to a point. How does that make sense?
Thanks for the help!
Edit: Here's what I did so far. Since $dg(X)=X(g)$, we must have for all $x \in M_1$, $dF(V_x)(f)=V_x(f \circ F)=V(f\circ F)(x)$. On the other hand, by defintition, we have $V_x(F^*f)=V(f \circ F)(x)$. Does this sound reasonable?
Perhaps it will be useful to look at the situation in local coordinates: fix local coordinates $x=(x^1,\dots,x^n)$ on $M_1$ and $y=(y^1,\dots,y^m)$ on $M_2$. The map $F$ will locally be written as $$F(x)=(F^1(x),\dots,F^m(x))$$ and for a vector field $V=v^a\partial_{x^a}$ we have \begin{equation} \begin{split} V_x(F^*f)=&v^a\partial_{x^a}\left(f(F^1(x),\dots,F^m(x))\right)=v^a(\partial_{y^b}f)(F(x))\partial_{x^a}F^b(x)=\\ =&\mathrm{d}f_{F(x)}\left((DF)_x(V)\right). \end{split} \end{equation} This last expression is just a different way to write the derivative of $f$ at the point $F(x)$ along the tangent vector $DF_x(V)\in T_{F(x)}M_2$, so the computation shows the formula you required.
P.s. to be honest, this formula is what I usually consider to be the definition of $DF:TM_1\to TM_2$. Perhaps you use a different definition?