Directional derivatives at $P$ are all derivations at $P$ Hi i am reading An introduction to Manifolds by Loring. And have one doubt under section 2.3 Derivations at a point. It is written that

we know that directional derivatives at $p$ are all derivations at $p$, so there is a map $$\phi:T_{p}\mathbb{R}^n\to D_{p}\mathbb{R}^n\\
v\mapsto D_v=\Sigma v^i \left.\frac{\partial}{\partial x^i}\right|_p$$

My doubt is that how is there a map as defined above. For example i know that $D_v f=\Sigma v^i \left.\frac{\partial f}{\partial x^i}\right|_p$ but how does this tell us that there is a map $\phi$ from $T_{p}\mathbb{R}^n\to D_{p}\mathbb{R}^n$ ? Note that $D_{p}\mathbb{R}^n$
is the set of all derivations at $p$ and $T_{p}\mathbb{R}^n$ is the set of all tangent vectors at point $p$. For reference i am attaching the screenshot of the portion where i have highlighted the part where this is mentioned. And what does it mean to have a map $\phi$ from $T_{p}\mathbb{R}^n$ to $D_{p}\mathbb{R}^n$?

 A: It is explained in the beginning of the paragraph "Tangent Vectors in $\mathbb R^n$ as Derivations / The Directional Derivative". Tu introduces the tangent space $T_p(\mathbb R^n)$ at $p$ in $\mathbb R^n$ as the vector space of all arrows emanating from $p$. Formally it is the set of all pairs $(p,q)$ with $q \in \mathbb R^n$, that is $T_p(\mathbb R^n) = \{p\} \times \mathbb R^n$. It can be identified with $\mathbb R^n$: To any $v \in \mathbb R^n$ we can associate the arrow from $p$ to $p+v$. Conversely, to any arrow from $p$ to $q$ we can associate the vector $q - p \in \mathbb R^n$.
For each $v \in \mathbb R^n$ we have the usual directional derivative $D_v = D_v \mid_p$ at $p$. This is the map which associates to any smooth $f : U \to \mathbb R$ defined on an open  neigborhood $U$ of $p$
$$D_v(f) = \lim_{t \to 0}\frac{f(p +tv)-f(p)}{t} .$$
Tu formalizes this and gets a map
$$D_v : C^\infty_p \to \mathbb R$$
defined on the set $C^\infty_p$ of germs of smooth functions $p$. This set is an $\mathbb R$-algebra (i.e. a vector space with a compatible multiplication of its elements). Tu shows that each $D_v$ is a derivation and thus gets his linear map
$$\phi : T_p(\mathbb R^n) \to \mathcal D_p(\mathbb R^n), \phi(v) = D_v.$$
The formula $D_v =  \sum v^i \dfrac{\partial}{\partial x_i} \mid_p$ follows from the fact that $D_{e_i} = \dfrac{\partial}{\partial x_i} \mid_p$.
