If I understand correctly the following is true:
- $T_{p}(\mathcal{M})$ is a vector space of linear derivations $X_{p}\in T_{p}(\mathcal{M})$, $X_{p}:C^{\infty}(\mathcal{M}) \rightarrow \mathbb{R}$
- Given a smooth map $F$ between manifolds $\mathcal{M}$,$\mathcal{N}$, $F:\mathcal{M} \rightarrow \mathcal{N}$ the pushforward of $F$, $F_{*}$ is defined as:
$$ F_{*}:T_{p}(\mathcal{M}) \rightarrow T_{F(p)}(\mathcal{M})\; ; \; (F_{*}(X))(f)=X(f\circ F)$$
Where $ X \in T_{p}(\mathcal{M}), f \in C^{\infty}(\mathcal{M})$
However this definition makes no sense to me. By definition: $f: \mathcal{M}\rightarrow \mathbb{R}$, so how is the composition $f\circ F$ defined? For the action of $X$ to be defined, we need $f\circ F \in C^{\infty}(\mathcal{M})$,which doesn't seem possible.
So surely we need $f \in C^{\infty}(\mathcal{N})$?, i.e my book has a printing error?
First of all, the pushforward $F_*$ has to be defined on $F_∗:T_p(M)→T_{F(p)}(N)$ since $F(p) \in N$. So clearly $f$ needs to be in $ C^{\infty}(N)$. You do that that you use a map $M \rightarrow N$ to construct a map on $T_pM \rightarrow T_qN.$ Your pushforward $F_*$ pushes a derivation on $T_p(M)$ forward to a derivation on $T_p(N)$.
So there might be one or two typos in your book. For nicer references, take Lee's Introduction to Smooth Manifolds or Tu's book on smooth manifolds.