Prove that tangent spaces, modeled as equivalence classes of curves, are vector spaces

Tangent vectors a $p \in M$ are equivalence classes of smooth curves $\sigma : (-\epsilon,\epsilon) \to M$ such that $\sigma(0) = p$ ("smooth curves in $M$ through $p$"). Here $\epsilon = \epsilon (\sigma)$ is a parameter which may vary from curve to curve. The equivalence relation is given by $\sigma_1 \sim \sigma_2$ if $(\phi \sigma_1)'(0) = (\phi \sigma_2)'(0)$ for some chart $\phi$ around $p$. It is easy to verify that $\sigma_1 \sim \sigma_2$ iff $(\phi \sigma_1)'(0) = (\phi \sigma_2)'(0)$ for all charts $\phi$ around $p$.

Given a smooth curve $\sigma : (-\epsilon,\epsilon) \to M$ through $p$, you can of course define $r \cdot \sigma : (-\epsilon/\lvert r \rvert,\epsilon/\lvert r \rvert) \to M, (r \cdot \sigma)(t) = \sigma (rt)$. Unfortunately there is no similar definition of $\sigma_1 + \sigma_2$ for curves $\sigma_i$ in $M$ trough $p$. You try to add them via the definition $$\sigma_1 + \sigma_2 = \phi^{-1}(\phi\sigma_1 + \phi \sigma_2).$$ This exploits the fact that the chart $\phi : U \to V \subset \mathbb R^n$ take values in $\mathbb R^n$, but in general it does not work because you cannot be sure that $\phi\sigma_1(t) + \phi \sigma_2(t) \in V$ for $\lvert t \rvert$ sufficiently small. Not even $\phi\sigma_1(0) + \phi \sigma_2(0) = \phi(p) + \phi(p) = 2\phi(p)$ is in general contained in $V$.

The solution is to consider only charts such that $\phi(p) = 0$. This can always be achieved if we replace an arbitrary chart $\phi$ by $T\phi$ where $T$ is the translation by $-\phi(p)$. The same holds for your definition of $r \cdot \sigma$.

Doing so, you will see that you get in fact the structure of a vector space on $T_p M$. Formally I suggest to proceed as follows:

1. Show that $\phi_* : T_pM \to T_0V, \phi_*([\sigma]) = [\phi\sigma]$, is a bijection.

2. Show that $T_0V$ becomes a vector space via $[\tau_1] + [\tau_2] = [\tau_1 + \tau_2]$ and $[r \cdot \tau] = [r \cdot \tau]$, where $(\tau_1 + \tau_2(t) = \tau_1(t)+ \tau_2(t)$ and $(r \cdot \tau)(t) = r \cdot \tau(t)$. Note that there always exist a maximal interval on which $\tau_1(t)+ \tau_2(t) \in V$ and $r \cdot \tau(t) \in V$; we take these intervals as the domains of $\tau_1 + \tau_2$ and $r \cdot \tau$. It is then easy to see that the map $\mathbb R^n \to T_0V, v \mapsto \tau_v$ with $\tau_v(t) = tv$, gives an isomorphism of vector spaces whichs shows that $\dim T_0V = n$.

3. Observe that $\phi_*$ induces a unique structure of a vector space on $T_pM$ such that $\phi_*$ becomes an isomorphism of vector spaces.

4. At first glance it seems that the vector space structure on $T_pM$ depends on the choice of $\phi$. The final step will therefore be to prove that any two charts $\phi_1, \phi_2$ around $p$ with $\phi_i(p) = 0$ produce the same vector space structure on $T_pM$.