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 $\chi^{\prime}$. 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).
For $p \in M$, 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 $\chi$ a chart around $p$. Let $\gamma^\chi := \chi \circ \gamma$ 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. I already proved that
Theorem 1 Given a smooth map $f: M \rightarrow N$ between two smooth manifolds, and given $p \in M$, there exists a unique linear map
$$
\mathrm d f_p: T_p M \rightarrow T_{f(p)} N, \quad v \mapsto \mathrm d f_p(v)
$$
with the property that, for any chart $\chi$ of $M$ around $p$ and $\chi^{\prime}$ of $N$ around $f(p)$, one has:
$$
\left( \mathrm d f_p(v)\right)^{\chi^{\prime}} = \mathrm d f_{\chi, \chi^{\prime}} (\chi(p)) [v^\chi] \quad \forall v \in T_p M . \qquad \qquad (1.3)
$$
Here $f_{\chi, \chi'} := \chi' \circ f \circ \chi^{-1}$ with domain $D(f_{\chi, \chi^{\prime}}) = \chi(U \cap f^{-1} (U')$ is the representation of $f$ w.r.t. $\chi, \chi'$.
Now I would like to prove another characterization of the differential of a smooth map, i.e.,
Theorem 2 Given a smooth map $f: M \rightarrow N$ between two smooth manifolds, and given $p \in M$, there exists a unique linear map
$$
\mathrm d f_p: T_p M \rightarrow T_{f(p)} N, \quad v \mapsto \mathrm d f_p(v)
$$
with the property that
$$
\mathrm d f_p \left(\frac{\mathrm d \gamma}{\mathrm d t}(0)\right) = \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0) \quad \forall \gamma \in \operatorname{Curve}_p (M). \qquad \qquad (1.4)
$$
With the help of Theorem 1, I'm able to show the existence of such map.
Could you elaborate why such map is unique?
My attempt Let $F$ be the linear map given by Theorem 1. Then we can verify that $F$ satisfies (1.4). Fix a chart $\chi$ of $M$ around $p$ and a chart $\chi^{\prime}$ of $N$ around $f(p)$. By (1.3) and chain rule,
$$
\begin{align}
\left [ F \left ( \frac{\mathrm d \gamma}{\mathrm d t}(0) \right ) \right ]^{\chi'} &= \mathrm d f_{\chi, \chi^{\prime}} (\chi(p)) \left [ \frac{\mathrm d \gamma}{\mathrm d t}(0) \right ]^\chi \\
&= \mathrm d f_{\chi, \chi^{\prime}} (\chi(p)) \circ \mathrm d (\chi \circ \gamma)(0)[1] \\
&= \mathrm d (\chi' \circ f \circ \chi^{-1}) (\chi(p)) \circ \mathrm d (\chi \circ \gamma)(0)[1] \\
&= \mathrm d (\chi' \circ f \circ \chi^{-1} \circ \chi \circ \gamma) (0) [1] \\
&= \mathrm d (\chi' \circ f \circ \gamma) (0) [1] \\
&= \left ( \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0) \right )^{\chi'}.
\end{align}
$$
Update My proof of uniqueness is as follows. However, I'm happy to see other approaches.
Let $v \in T_pM$. There exists $\gamma \in \operatorname{Curve}_p (M)$ such that $v = \frac{\mathrm d \gamma}{\mathrm d t} (0)$. It follows that
$$
F (v) = \frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0).
$$
It suffices to prove that if $\gamma, \lambda \in \operatorname{Curve}_p (M)$ with $v = \frac{\mathrm d \gamma}{\mathrm d t} (0) = \frac{\mathrm d \lambda}{\mathrm d t} (0)$, then $\frac{\mathrm d (f \circ \gamma)}{\mathrm d t}(0) = \frac{\mathrm d (f \circ \lambda)}{\mathrm d t}(0)$. Fix a chart $\chi$ of $M$ around $p$ and $\chi^{\prime}$ of $N$ around $f(p)$. Then we have $\frac{\mathrm d (\chi \circ \gamma)}{\mathrm d t} (0) = \frac{\mathrm d (\chi \circ \lambda)}{\mathrm d t} (0)$. It suffices to show that $\frac{\mathrm d (\chi' \circ f \circ \gamma)}{\mathrm d t}(0) = \frac{\mathrm d (\chi' \circ f \circ \lambda)}{\mathrm d t}(0)$. Indeed,
$$
\begin{align}
\frac{\mathrm d (\chi' \circ f \circ \gamma)}{\mathrm d t}(0) &= \mathrm d (\chi' \circ f \circ \gamma)(0)[1] \\
&= \mathrm d (\chi' \circ f \circ \chi^{-1} \circ \chi \circ \gamma)(0)[1] \\
&= \mathrm d (\chi' \circ f \circ \chi^{-1}) (\chi (p)) \circ \mathrm d (\chi \circ \gamma)(0)[1].
\end{align}
$$
Similarly,
$$
\frac{\mathrm d (\chi' \circ f \circ \gamma)}{\mathrm d t}(0) = \mathrm d (\chi' \circ f \circ \chi^{-1}) (\chi (p)) \circ \mathrm d (\chi \circ \lambda)(0)[1].
$$
The proof is completed by
$$
\frac{\mathrm d (\chi \circ \gamma)}{\mathrm d t} (0) := \mathrm d (\chi \circ \gamma)(0)[1]
\quad \text{and} \quad
\frac{\mathrm d (\chi \circ \lambda)}{\mathrm d t} (0) := \mathrm d (\chi \circ \lambda)(0)[1].
$$
Best Answer
You do not need any calculation. You know that
$df_p : T_pM \to T_{f(p)}$ is a linear map satisfying $(1.4)$.
the function $\frac{d}{dt}(0) : \operatorname{Curve}_p (M) \to T_pM, \gamma \mapsto \frac{d\gamma}{dt}(0)$, is surjective.
$(1.4)$ means that you have a commutative diagram $\require{AMScd}$ \begin{CD} \operatorname{Curve}_p (M) @>{f_*}>> \operatorname{Curve}_{f(p)} (N) \\ @V{\frac{d}{dt}(0)}VV @VV{\frac{d}{dt}(0)}V \\ T_pM @>>{df_p}> T_{f(p)}N \end{CD} where $f_*(\gamma) = f \circ \gamma$. Now assume that $F : T_pM \to T_{f(p)}N$ is any function (you do not even need to assume that it is linear) which also fits into the above diagram (i.e. $F \circ \frac{d}{dt}(0) = \frac{d}{dt}(0) \circ f_*$). Then $$F \circ \frac{d}{dt}(0) = df_p \circ \frac{d}{dt}(0) .$$ The surjectivity of $\frac{d}{dt}(0)$ immediatele implies that $F = df_p$.
This is just a little set-theoretic lemma:
If $s : X \to Y$ is surjective and $g,h : Y \to Z$ are any two functions such that $g \circ s = h \circ s$, then $g = h$.
The proof is obvious: Let $y \in Y$. Choose $x \in X$ such that $s(x) = y$. Then $g(y) = (g \circ s) (x) = (h \circ s)(x) = h(y)$.