Suppose we have a curve $\alpha: \mathbb{R} \rightarrow \mathbb{R}^2$ through $p \in \mathbb{R}^2$ and $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is a smooth function. Then $f \circ \alpha: \mathbb{R} \rightarrow \mathbb{R}$ is a smooth function and we can differentiate it at $0$: $$ (f \circ \alpha)^{\prime}(0)=a_1^{\prime}(0) \frac{\partial f}{\partial x_1}(\underbrace{p}_{\alpha(0)})+a_2^{\prime}(0) \frac{\partial f}{\partial x_2}(\underbrace{p}_{\alpha(0)}) $$ by the Chain rule. We therefore have a map $f \mapsto(f \circ \alpha)^{\prime}(0)$ from functions to $\mathbb{R}$ given by $$ f \mapsto\left(\left.a_1^{\prime}(0) \frac{\partial}{\partial x_1}\right|_p+\left.a_2^{\prime}(0) \frac{\partial}{\partial x_2}\right|_p\right) f, $$ which is a differential operator acting on functions. If we think of $\left\{\left.\frac{\partial}{\partial x_1}\right|_p,\left.\frac{\partial}{\partial x_2}\right|_p\right\}$ as a basis for a $2$ dimensional vector space, then we identify this map is the tangent vector to $\alpha$ at $p$.
The good thing about this is that we can replace $\mathbb{R}^2$ by any manifold $M$, since $f \circ \alpha: \mathbb{R} \rightarrow \mathbb{R}$ can be differentiated, so this definition still works. Explicitly, if $\alpha: \mathbb{R} \rightarrow M$ is a curve through $p \in M$ and $f: M \rightarrow \mathbb{R}$ is a smooth function then we let $(U, \varphi)$ be a coordinate chart at $p$ and write $\varphi \circ \alpha(t)=$ $\left(a_1(t), \ldots, a_n(t)\right) \in \varphi(U) \subseteq \mathbb{R}^n$. Then $$ \begin{align} (f \circ \alpha)^{\prime}(0)&=\left.\frac{\mathrm{d}}{\mathrm{d} t}(f \circ \alpha)(t)\right|_{t=0}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d} t}\left(f \circ \varphi^{-1} \circ \varphi \circ \alpha\right)(t)\right|_{t=0}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d} t}\left(f \circ \varphi^{-1}\right)\left(a_1(t), \ldots, a_n(t)\right)\right|_{t=0}\tag1\\ &=\sum_{j=1}^n a_j^{\prime}(0) \frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial x_j}(\varphi(p))\tag2\\ &=\left(\left.\sum_{j=1}^n a_j^{\prime}(0) \frac{\partial}{\partial x_j}\right|_{\varphi(p)}\right)\left(f \circ \varphi^{-1}\right) \end{align} $$ Hence, using the $\left.\frac{\partial}{\partial x_j}\right|_{\varphi(p)}$ as a basis, we can identify the tangent vector to the curve $\varphi \circ \alpha$ in $\mathbb{R}^n$ at $\varphi(p)$ with the differential operator $\left.\sum_{j=1}^n a_j^{\prime}(0) \frac{\partial}{\partial x_j}\right|_{\varphi(p)}$ acting on the function $f \circ \varphi^{-1}$ (which is how we identify functions on $M$ locally with functions on $\mathbb{R}^n$ ).
In the first paragraph, considering $\left\{\left.\frac{\partial}{\partial x_1}\right|_p,\left.\frac{\partial}{\partial x_2}\right|_p\right\}$ as a basis make sense, as it can be thought of as the small increment/direction of each coordinate direction in $\mathbb R^2$. But that's not happening for second paragraph, specially $\frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial x_j}$. Because there is no coordinate system, as we are considering in the abstract manifold space $M$.
- Then how to interpret $\frac{\partial}{\partial x_j}$ here? And I am unable to understand how $(2)$ come from $(1)$?
- Why the summation follow the dimension of the chart? And what's that mean?
It will be a great help if anyone can help me to understand the definition. TIA
Updated
Definition 4.2.1: Given a $C^k$ manifold, $M$, of dimension $n$, for any $p \in M$, two $C^1$-curves, $\left.\gamma_1:\right]-\epsilon_1, \epsilon_1[\rightarrow M$ and $\left.\gamma_2:\right]-\epsilon_2, \epsilon_2\left[\rightarrow M\right.$, through $p$ (i.e., $\gamma_1(0)=\gamma_2(0)=p$ ) are equivalent iff there is some chart, $(U, \varphi)$, at $p$ so that $$ \left(\varphi \circ \gamma_1\right)^{\prime}(0)=\left(\varphi \circ \gamma_2\right)^{\prime}(0) . $$ Definition 4.2.2 (Tangent Vectors, Version 1): Given any $C^k$-manifold, $M$, of dimension $n$, with $k \geq 1$, for any $p \in M$, a tangent vector to $M$ at $p$ is any equivalence class of $C^1$-curves through $p$ on $M$, modulo the equivalence relation defined in Definition 4.2.1. The set of all tangent vectors at $p$ is denoted by $T_p(M)$.
One of the defects of the above definition of a tangent vector is that it has no clear relation to the $C^k$-differential structure of $M$. There is another way to define tangent vectors that reveals this connection more clearly. As a first step, consider the following: Let $(U, \varphi)$ be a chart at $p \in M$ (where $M$ is a $C^k$-manifold of dimension $n$, with $k \geq 1$ ) and let $x_i=p r_i \circ \varphi$ (where $pr_i: \mathbb R_n \rightarrow \mathbb R$, are defined by $pr_i(x_1, \cdots x_n) = x_i, 1\leq i\leq n)$, the $i$ th local coordinate $(1 \leq i \leq n)$. For any function, $f$, defined on $U \ni p$, set $$ \left(\frac{\partial}{\partial x_i}\right)_p f=\left.\frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial X_i}\right|_{\varphi(p)}, \quad 1 \leq i \leq n . $$ (Here, $\left.\left(\partial g / \partial X_i\right)\right|_y$ denotes the partial derivative of a function $g: \mathbb{R}^n \rightarrow \mathbb{R}$ with respect to the $i$ th coordinate, evaluated at $y$.)
Here the partial derivatives are defined as $$ \left(\frac{\partial}{\partial x_i}\right)_p f=\left.\frac{\partial\left(f \circ \varphi^{-1}\right)}{\partial X_i}\right|_{\varphi(p)}, \quad 1 \leq i \leq n . $$ which seems different from @J.V.Gaiter answer,
$$\frac{\partial}{\partial x_j}\vert_pf:= \frac{\partial (f\circ \phi)}{\partial x_j}(p)$$
I couldn't come up with their equivalence.
The $\frac{\partial}{\partial x_j}\vert_p$'s are objects that come with the coordinate charts we place on the manifold. We commit abuse of notation (or convenient identification) by saying $\frac{\partial}{\partial x_j}\vert_p\in T_pM$ defined by $\frac{\partial}{\partial x_j}\vert_pf:= \frac{\partial (f\circ \phi)}{\partial x_j}(p)$ for $\phi$ the chart for which $x_j$ is a coordinate.
If you think of the tangent vectors as "infinitesimal displacements" then the fact that for an $n$-dimensional manifold, $T_pM$ is $n$-dimensional should be manifest. If in a coordinate chart there are $n$-dimensions in which we can move, then the space of infinitesimal displacements should have that same dimension.
EDIT:
In order to get the fact that the space of tangent vectors is $n$-dimensional on an $n$-dimensional manifold, one needs to apply Taylor's theorem.
If $f\in C^\infty(M)$ and $\phi: U\to \mathbb{R}^n$ is a chart, then in a small neighborhood of $p\in U$ we can write $f(x+p)=f(p)+df(x)+\sum_{i,j} A_{ij}(x)x^ix^j$ for some smooth functions $A_{ij}(x)$. $df$ is a linear map $\mathbb{R}^n\to \mathbb{R}$. The Leibniz rule tells us that for a derivation based at $p$, $\delta: C^\infty(M)\to \mathbb{R}$, we have $\delta(f(x+p))=\delta(f(p))+\delta(Df)(p)+\delta(\sum_{ij}A_{ij}(x)x^{i}x^j)=\delta(Df)(p)$. This means that $\delta$ only depends on the value of $Df(p)$. As $df(p)$ is an element of $(\mathbb{R}^n)^*$ it lies in an $n$-dimensional vector space and we can construct functions $f\in C^\infty(M)$ with $Df=\xi$ for $\xi\in (\mathbb{R}^n)^* $with respect to any given compatible coordinate chart using bump functions. This means that the dimension of the space of derivations at $p$, $\mathrm{Der}_p(M)$ is at most $n$. We can show that the space is $n$ dimensional by using the coordinate curves, i.e. we can find $n$ linearly independent elements of $\mathrm{Der}_p(M)$ given by the maps $f\mapsto \frac{d}{dt}\vert_{t=0}f(p+tx^i)$.
(Note, here I am abusing notation and identifying $f:M\to \mathbb{R}$ with $f\circ \phi: U\to \mathbb{R}$)