I am a bit stuck about figuring out that the result of a differential map $f_{\ast p}$ defined on a tangent space is a derivation in the "other" tangent space.
Suppose $f \colon N \to M$ is a map between a $n$-dimensional manifold $N$ and a $m$-dimensional manifold $M$. Define $f_{\ast p}$ as the map
$v_p \in T_p(N) \mapsto f_{\ast p}(v_p) \in T_{f(P)}(M)$ such that $f_{\ast p}(v_p)(g) = v_p(g \circ f)$
where $g$ is a differentiable function from $M$ to $\mathbb{R}$. Let $(U,u)$ be a chart around $P \in N$ and let $(V,v)$ a chart around $f(p) \in M$. I am a bit skeptical because, following the definition, it seems to have defined just another derivation on $T_p(N)$ (well, actually it is), i. e.
$\frac{\partial}{\partial x_i}(g\circ f\circ u^{-1})(u(P))$.
My question is: is $f_{\ast p}(v_p)$ a derivation on $T_p(M)$ too because of the relation
$g\circ f\circ u^{-1} = (g \circ v^{-1})(v\circ f\circ u^{-1})$ ?
If that is the case, then it seems to work because we have a derivation that sends a point in $\mathbb{R}^m$ (that is $v(P)$) to a point of $\mathbb{R}$ but I'm not so sure. Thanks.