I'm reading about tangent vectors from this lecture note.
Given an $m$-dimensional manifold $M$ and $p \in M$, a tangent vector of $M$ at $p$ is any function $$ v:\{\text{charts of } M \text { around } p\} \rightarrow \mathbb{R}^m, \quad \chi \mapsto v^\chi, $$ with the property that, for any two charts $\chi$ and $\chi^{\prime}$ around $p$ one has $$ v^{\chi^{\prime}}=(\mathrm d c)_{\chi(p)}\left(v^\chi\right), \qquad (1.2) $$ where $c=c_{\chi, \chi^{\prime}}$ is the change of coordinates from $\chi$ to $\chi^{\prime}$. We denote by $T_p M$ the vector space of all such tangent vectors of $M$ at $p$ (a vector space using the vector space structure on $\mathbb{R}^m$, i.e. $(v+w)^\chi:=v^\chi+w^\chi$, etc).
LEMMA 4.2. If $\chi_0$ is a chart of $M$ around $p$, then $$ T_p M \rightarrow \mathbb{R}^m, v \mapsto v^{\chi_0} $$ is an isomorphism of vector spaces.
Proof. As we have already remarked, given $v_0 \in \mathbb{R}^m$ then the $v \in T_p M$ such that $v^{\chi_0}=v_0$ is unique: it must be given by the formula $$ v^\chi=\left(\mathrm d c_{\chi_0, \chi}\right)_{\chi_0(p)}\left(v_0\right), $$ for any other chart $\chi$ around $p$. Strictly speaking we still prove (1.2) but that follows from the chain rule and the remark that $c_{\chi_0, \chi^{\prime}}=c_{\chi, \chi^{\prime}} \circ c_{\chi_0, \chi}$.
My understanding Let $u, v \in T_p M$ such that $u^\chi = v^\chi$. Let $(\chi', U')$ be a chart of $M$ around $p$ and $c:=\chi' \circ \chi^{-1}$. Then $$ v^{\chi^{\prime}} = (\mathrm d c)_{\chi(p)}\left(v^\chi\right) = (\mathrm d c)_{\chi(p)}\left(u^\chi\right) = u^{\chi'}. $$
Because $(\chi', U')$ is arbitrary, we get $u=v$. Then the map is injective.
Could you explain how the author shows that the map is surjective?
Let $\mathfrak C_p(M)$ denote the set of charts of $M$ around $p$. Then $T_pM$ is defined as the set of all functions $v : \mathfrak C_p(M)\to \mathbb R^n$ such that $(1.2)$ is satisfied. Note that the author of your notes prefers to write $v^\chi$ instead of $v(\chi)$.
The claim is that for each $\chi_0 \in \mathfrak C_p(M)$ we get an isomorphism
$$\chi_0^* : T_pM \to \mathbb R^n, \chi_0^*(v) = v^{\chi_0} . $$
By definition of the vector space structure on $T_pM$ it is obviuosly a linear map. It remains to show that it is bijective.
For $\chi,\chi' \in \mathfrak C_p(M)$ let $c_{\chi,\chi'} = \chi' \circ \chi^{-1}$ denote the coordinate change from $\chi$ to $\chi'$. Given $\xi \in \mathbb R^n$, define $$\overline{\xi} : \mathfrak C_p(M) \to \mathbb R^n,\overline{\xi}(\chi) = \overline{\xi}^\chi = (dc_{\chi_0, \chi})_{\chi_0(p)} (\xi). $$ This function satisfies $(1.2)$: Let $\chi,\chi' \in \mathfrak C_p(M)$. Then $$\overline{\xi}^{\chi'} = (dc_{\chi_0,\chi'})_{\chi_0(p)} (\xi) = (dc_{\chi,\chi'})_{\chi(p)}(dc_{\chi_0,\chi})_{\chi_0(p)} (\xi) = (dc_{\chi,\chi'})_{\chi(p)}(\overline{\xi}^{\chi}) .$$ This means $\overline{\xi} \in T_pM$. We have $c_{\chi_0,\chi_0} = id$, thus $(dc_{\chi_0, \chi_0})_{\chi_0(p)} = id$ and by definition we get $\chi_0^*(\overline{\xi}) = \overline{\xi}^{\chi_0} = \xi$.
Since $\chi_0^*$ is linear, it suffices to show that $v^{\chi_0} = \chi_0^*(v) = 0$ implies $v = 0$ (i.e. $v^\chi = 0$ for all $\chi$). But this follows immediately from $(1.2)$.