Is each tangent vector of $M$ at $p$ induced by a curve in $\operatorname{Curve}_p (M)$? 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
$$
v^{\chi^{\prime}} = \mathrm d c_{\chi, \chi^{\prime}} (\chi(p)) [v^\chi], \qquad \qquad (1.2)
$$
where $c_{\chi, \chi^{\prime}} :=\chi' \circ \chi^{-1}$ is the change of coordinates from $\chi$ to $\chi^{\prime}$. Here $\mathrm d c_{\chi, \chi^{\prime}} (\chi(p))$ is the derivative of $c_{\chi, \chi^{\prime}}$ at $\chi(p)$, and $\mathrm d c_{\chi, \chi^{\prime}} (\chi(p)) [v^\chi]$ its value at $v^\chi$. 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). Let $p \in M$, we define
$$
\operatorname{Curve}_p (M) := \{\gamma: (-\epsilon, \epsilon) \to M \text{ smooth such that } \epsilon>0, \gamma (0)=p\}.
$$
Let $\gamma \in \operatorname{Curve}_p (M)$ and $(\chi, U)$ a chart around $p$. Let $\gamma^\chi := \chi \circ \gamma : (-\epsilon, \epsilon) \to \mathbb R^m$ be the representation of $\gamma$ w.r.t. $\chi$. Consider the operation
$$
\frac{\mathrm d \gamma}{\mathrm d t} (0):\{\text {charts of } M \text { around } p\} \rightarrow \mathbb{R}^m, \quad \chi \mapsto \frac{\mathrm d \gamma^\chi}{\mathrm d t}(0).
$$
Then $\frac{\mathrm d \gamma}{\mathrm d t} (0) \in T_pM$ by chain rule. This means each curve in $\operatorname{Curve}_p (M)$ gives rise to an element in $T_pM$. I would like to ask if the reverse is true, i.e.,

Is each tangent vector of $M$ at $p$ induced by a curve in $\operatorname{Curve}_p (M)$?


Update If the answer is affirmative, then the uniqueness part in this thread follows easily.
 A: Yes. If we fix a chart $\chi_0$, for any $v_0\in\mathbb R^n$, we can define a tangent vector $\chi\mapsto d(\chi\circ \chi_0^{-1})(v_0)$. Indeed, we have $$v^{\chi}=d(\chi\circ\chi_0^{-1})(v_0)=d(\chi\circ\chi'^{-1}\circ\chi'\circ\chi_0^{-1})(v_0)=d(\chi\circ\chi'^{-1})\circ d(\chi'\circ\chi_0^{-1})(v_0)=d(\chi\circ\chi'^{-1})v^{\chi'}$$
On the other hand, given a tangent vector $v$, we must have it's uniqiely determined by $v^{\chi_0}$: Just let $\chi=\chi_0$ in $(1.2)$, $v^{\chi'}$ is determined by $v^{\chi_0}$ for any $\chi'$. That is, fix a chart $\chi_0$, the tagent vectors at $p$ can be identified with $\mathbb R^n$.
Now it suffices to show that given a point $p$, chart $\chi$ and vector $(a_1, \cdots, a_n)$, there is a curve $\gamma: (-\epsilon, \epsilon)\rightarrow M$ such that $\gamma(0)=p$, $\frac{d(\chi\circ\gamma)}{dt}(0)=(a_1, \cdots, a_n)$.
But this is very simple: Let $\gamma(t) = \chi^{-1}(\chi(p)+ t(a_1, a_2, \cdots, a_n))$ that clearly satisfies $\gamma(0)=p$, and $$(\chi^{-1}\circ\gamma)(t) = \chi(p) + t(a_1, a_2, \cdots, a_n)$$
$$\frac{d(\chi^{-1}\circ\gamma)(t)}{dt}=(a_1, a_2, \cdots, a_n)$$
Beginners like to study differential geometry as we did real analysis, trying hard to follow the texts rigorously, but maybe it's more important to keep track of the essential ideas like we did elementary calculus, and rigor will come naturally. In this case, intuitively, a tangent vector is just the direction of any curve that goes, and it can be made explicit in terms of coordinates once a chart is fixed, but the coordinates would change with respect to different charts. Different rules of change would give different types of "tensors", and differential geometry is more or less about the study of the properties that don't change under different charts.
A: Yes, assuming smooth manifolds and curves. In fact another definition of $T_pM$ is the following
$$T_pM = \text{Curve}_p(M)/\sim$$
Where the equivalence relation is defined as the following let $\phi$ be a chart of a neighborhood around $p$. Then for $\gamma, \eta \in \text{Curve}_p(M) $
$$\gamma \sim \eta \hspace{5mm} \iff\hspace{5mm} \frac{d(\phi\circ\gamma)}{dt}(0)=\frac{d(\phi\circ\eta)}{dt}(0)$$
So this means that the tanget space is essentialy the set of equivalence classes of curves that pass through $p$ which have the same "velocity" when passing through the point $p$ and that is thought of as the tangent vector at $p$. This means that there are many curves that represent a single vector in $T_pM$.
