Show that sum of directional derivatives at a point on a manifold doesn't depend on the representation of the curve If $ v_1 , v_2 \in T_{p} \mathcal M$ . $p$ is a point on a manifold $\mathcal M$,( $p \in \mathcal M$),
The tangent space $T_p \mathcal M$ is given a vector space structure by defining
\begin{align}
 v_1 + v_2 := [\phi^{-1} \circ ( \phi \circ \sigma_1 + \phi \circ \sigma_1 )]
\end{align}
Here $\phi$ is the map that maps from the open set of $p \in \mathcal M$ to $\mathbb{R}^m$.
$\phi(0) = \mathbf{0}$
And $\sigma_1,\sigma_2 : (-\epsilon,\epsilon) \to \mathcal M$ are curves. Such that
$v_1 = [\sigma_1]$ and $v_2 = [\sigma_2]$
I want to prove that this definition is independent of charts.
To do that

*

*I use the definition of directional derivatives
\begin{align}
 v(f) := \left. \frac{d(f\circ \sigma)}{dt}\right|_{t=0} = (f\circ \sigma)'(0) \quad v = [\sigma]
\end{align}
This is independet of the representation of $[\sigma]$, (shown here)

*I use chain rule as per the multi variable calculus

*Linearity of derivarives is also used

This is my attempt of the proof
\begin{align}
 (v_1 + v_2)(f) &= (f \circ \phi^{-1} \circ ( \phi \circ \sigma_1 + \phi \circ \sigma_2 ))'(0) 
\tag{1}\label{1} \\
&=  (f \circ \phi^{-1} )'\left(( \phi \circ \sigma_1 + \phi \circ \sigma_2 )(0)\right)\circ( \phi \circ \sigma_1 + \phi \circ \sigma_1 )'(0) \tag{2}\label{2} \\
&=(f \circ \phi^{-1} )'\left(0\right)\circ( \phi \circ \sigma_1 + \phi \circ \sigma_2 )'(0) \tag{3}\label{3} \\
&=(f \circ \phi^{-1} )'\left(0\right)\circ( \phi \circ \sigma_1  )'(0) +
(f \circ \phi^{-1} )'\left(0\right)\circ(\phi \circ \sigma_2 )'(0)  \tag{4}\label{4}
\\
&=(f \circ\sigma_1  )'(0) +(f \circ\sigma_2 )'(0)   \tag{5}\label{5}
\\
&= v_1(f) + v_2(f)\tag{6}\label{6}
\end{align}

*

*Writing the directional derivative along $ v_1 + v_2$

*Using the chain rule $(f \circ g )'(x) = f'(g(x))\circ g'(x)$

*Using ,$ \sigma_1(0) = \sigma_2(0) = p$, and $\phi(p) = \mathbf 0$

*Using the linearity of derivatives $ f'(0)\circ (g_1 + g_2)(0) = f'(0) \circ g_1(0) + f'(0) \circ g_2(0)$

*Reverse usage of chain rule and replacing $\phi^{-1} \circ \phi$ as $1$

*Recognizing the forms of directional derivatives

In the last step, each term in the expression is independent of the representation of the curve and also independent of the charts around $p \in \mathcal M$ for any arbitrary $f \in C^{\infty} (\mathcal M)$, so
$v_1 + v_2$ doesn't depend on the representation of the curves $\sigma_1$ or $\sigma_2$, any representation from $[\sigma_1]$ or $[\sigma_2]$ will work.
$$\tag*{$\blacksquare$}$$
Is this the right way of doing it? I am unsure about the linearity of the derivative \eqref{4}.
P.S: This proof is essential to show that the tangent space is a vector space.
I am following this book,

Isham, Chris J., Modern differential
geometry for physicists., World Scientific Lecture Notes in Physics.
61. Singapore: World Scientific. xiii, 289 p. (1999). ZBL0931.53002.

 A: I think your work shows that the directional derivative induced by the curve $\phi^{-1}\circ(\phi\circ \sigma_1+\phi\circ\sigma_2)$ does not depend of your choice of $\phi$, nor representatives of the equivalence classes of $\sigma_1$ and $\sigma_2$ (with $\sigma_1\sim\sigma_2$ iff there is a chart $(\phi,U)$ in p with $(\phi\circ\sigma_1)'(0)=(\phi\circ\sigma_2)'(0)$ $(\ast)$).
We know that tangent vectors and directional derivatives are the same through the linear isomorphism
$$v=[\sigma]\mapsto(f\mapsto (f\circ\sigma)'(0)),$$
but this latter already suppose we know that there is a vector space structure on the tangent vectors defined as equivalence classes of curves. So what we actually have to prove is that
$$\phi^{-1}\circ(\phi\circ \sigma_1+\phi\circ\sigma_2)\sim\psi^{-1}\circ(\psi\circ \tilde\sigma_1+\psi\circ\tilde\sigma_2)$$
for another $\psi:M\to\mathbb{R}^m$ s.t. $\psi(p)=0$, and $\sigma_1\sim\tilde\sigma_1,\sigma_2\sim\tilde\sigma_2$. Happily, the proof looks like a lot the one you already produced:
Choose $(\phi,U)$ as the map we will use to check $(\ast)$. The derivative of the first curve is
$$(\phi\circ\phi^{-1}\circ(\phi\circ\sigma_1+\phi\circ\sigma_2))'(0)=(\phi\circ\sigma_1)'(0)+(\phi\circ\sigma_2)'(0).$$
The linearity of the derivative is indeed checked here: for functions $F,G:\mathbb{R}\to\mathbb{R}^m$, we can write $F(t)=(F_1(t),\dots,F_m(t)),G(t)=(G_1(t),\dots,G_m(t))$ and since $(F+G)(t)$ is no more than $(F_1(t)+G_1(t),\dots,F_m(t)+G_m(t))$, the linearity is the one for functions from $\mathbb{R}$ to $\mathbb{R}$.
The derivative of the second curve is
\begin{align*}
&(\phi\circ\psi^{-1}\circ(\psi\circ \tilde\sigma_1+\psi\circ\tilde\sigma_2))'(0)\\
=\,&D(\phi\circ\psi^{-1})_0\left((\psi\circ\tilde\sigma_1)'(0)+(\psi\circ\tilde\sigma_2)'(0)\right)\\
=\,&D(\phi\circ\psi^{-1})_0\left((\psi\circ\tilde\sigma_1)'(0)\right)+D(\phi\circ\psi^{-1})_0\left((\psi\circ\tilde\sigma_2)'(0)\right)\\
=\,&(\phi\circ\tilde\sigma_1)'(0)+(\phi\circ\tilde\sigma_2)'(0)\\
=\,&(\phi\circ\sigma_1)'(0)+(\phi\circ\sigma_2)'(0),
\end{align*}
this last equality coming from $\sigma_i\sim\tilde\sigma_i$, $i=1,2$.
