Let $M$ be a smooth Manifold.
Let $p \in M$.
Let $U$ be an open neighborhood of $p$.
We denote by $C^{\infty}(U)$ the set of real valued smooth functions on $U$.
Let $\Lambda_p = \bigcup C^{\infty}(U)$, where $U$ runs through all open neighborhoods of $p$.
Let $f, g \in \Lambda_p$.
Suppose $f \in C^{\infty}(U)$ and $g \in C^{\infty}(V)$.
If there exists an open neighborhood $W$ of $p$ such that $W \subset U \cap V$ and $f|W = g|W$,
we say $f$ and $g$ are equivalent.
This is an equivalence relation on $\Lambda_p$.
We denote by $\mathcal{O}_p$ the set of equivalence classes on $\Lambda_p$.
Clearly $\mathcal{O}_p$ is an $\mathbb{R}$-algebra.
Let $f \in C^{\infty}(U)$, where $U$ is an open neighborhodd of $p$. We denote by $[f]$ the equivalence class containing $f$.
A derivation of $\mathcal{O}_p$ is a linear map $D\colon \mathcal{O}_p \rightarrow \mathbb{R}$ such that
$D(fg) = D(f)g(p) + f(p)D(g)$ for $f, g \in \mathcal{O}_p$.
The set $T_p(M)$ of derivations of $\mathcal{O}_p$ is a vector space over $\mathbb{R}$ and is called the tangent space at $p$.
Let $\epsilon > 0$ be a positive real number.
We denote by $\Gamma_p(\epsilon)$ the set of smooth curves $\gamma \colon (-\epsilon,\epsilon) \rightarrow M$ such that $\gamma(0) = p$.
Let $\Gamma_p = \bigcup_{\epsilon>0} \Gamma_p(\epsilon)$.
Let $(U, \phi)$ be a chart such that $p \in U$.
Let $\gamma_1, \gamma_2 \in \Gamma_p$.
Then $\gamma_1$ and $\gamma_2$ are called equivalent at $0$ if $(\phi\circ\gamma_1)'(0) = (\phi\circ\gamma_2)'(0)$. This definition does not depend on the choice of the chart $(U, \phi)$.
This defines an equivalence relation on $\Gamma_p(M)$.
Let $S_p(M)$ be the set of equivalence classes on $\Gamma_p(M)$.
For $\gamma \in \Gamma_p(M)$, we denote by $[\gamma]$ the equivalence class containing $\gamma$.
We will define a map $\Phi\colon S_p(M) \rightarrow T_p(M)$.
Let $c \in S_p(M)$.
Choose $\gamma \in \Gamma_p(M)$ such that $c = [\gamma]$.
Let $f \in C^{\infty}(U)$, where $U$ is an open neighborhood of $p$.
We write $D_c([f]) = (f\circ\gamma)'(0)$ for $f \in C^{\infty}(U)$.
Clearly $D_c$ is well defined and does not depend on the choice of $\gamma$.
Clearly $D_c \in T_p(M)$.
Hence we get a map $\Phi\colon S_p(M) \rightarrow T_p(M)$ such that $\Phi(c) = D_c$.
We claim that $\Phi$ is bijective.
Let $c, e \in S_p(M)$.
Suppose $D_c = D_e$.
Suppose $c = [\gamma]$ and $e = [\lambda]$.
Let $(U, \phi)$ be a chart such that $p \in U$.
Let $\pi_i:\mathbb{R}^n \rightarrow \mathbb{R}$ be the $i$-th projection map: $\pi_i(x_1,\dots,x_n) = x_i$.
We denote by $\phi^i$ by $\pi_i\circ\phi$.
Since $D_c([\phi^i]) = D_e([\phi^i])$, $(\phi^i\circ\gamma)'(0) = (\phi^i\circ\lambda)'(0)$.
Hence $(\phi\circ\gamma)'(0) = (\phi\circ\lambda)'(0)$.
Hence $\gamma$ and $\lambda$ is equivalent.
Thus $\Phi$ is injective.
Let $D \in T_p(M)$.
Let $(U, \phi)$ be a chart such that $p \in U$.
We assume that $\phi(p) = 0$.
We define $\phi^i$ for $i = 1,\dots,n$ as above.
Let $D([\phi^i]) = a_i$ for $i = 1,\dots,n$.
There exists $\epsilon > 0$ such that $(a_1t,\dots,a_nt) \in \phi(U)$ for every $t \in (-\epsilon, \epsilon)$.
Let $\gamma(t) = \phi^{-1}(a_1t,\dots,a_nt)$ for $t \in (-\epsilon, \epsilon)$.
Then it's easy to see that $\Phi([\gamma]) = D$.
Hence $\Phi$ is surjective and we are done.
An open map is just a function that maps an open set to an open set, whereas a continuous map satisfies that the preimage of an open set is an open set. Since a diffeomorphism $f$ is a continuous bijection, $f^{-1}$ is continuous. It follows that $f$ is an open map.
Since every homeomorphism is a continuous bijection, it is always an open map.
Best Answer
A smooth curve is (typically) a smooth map $f:[0,1]\to U$. And so if $\varphi:U\to V$ is a diffeomorphism then our smooth curve induces a smooth curve $g:I\to V$ given by the composition $g:=\varphi\circ f$. And so $f\mapsto \varphi\circ f$ is a function between sets of smooth curves with the inverse $g\mapsto \varphi^{-1}\circ g$. That's most likely what the author meant.