I am currently working with the definition of tangent vectors being equivalence classes of curves. So $v =[\gamma]$ and $w=[\sigma]$ where $v,w$ are the vectors. I want to prove that the sum of this two equivalence class is also a equivalence class to give it the vector space structure. We define the sum as:
$v+w= [\phi^{-1} \circ (\phi \circ \gamma + \phi \circ \sigma)]$
where $\phi$ is a chart. I understand that the curves can not be added when they belong to the manifold, that's why we first map then into the reals where that can be performed. So now that the sum is defined, how can I prove that is another class at the point $p$ on $M$. Is it enough to make sure that the map passes by $p$? The same questions goes for the multiplication by the scalar. Thanks
Having read your post more carefully, here's a one-sentence summary of your mistake: you're trying to add (and scalar multiply) the curves in $\Bbb{R}^n$, rather than their velocities. As you observed, adding the curves messes up things with the base points.
As a set, we have $T_pM$ is the set of equivalence classes of smooth curves, $[\gamma]$, where $\gamma$ is defined on some open interval containing $0$ such that $\gamma(0)=p$. Now, for any chart $(U,\phi)$ about the point $p$, consider the function $F_{\phi,p}:T_pM \to \Bbb{R}^n$ defined as \begin{align} F_{\phi,p}([\gamma]):= (\phi\circ \gamma)'(0). \end{align} This function is well-defined because of how the equivalence relation is defined. Notice the intuitive meaning: $\gamma$ is a curve with values in the manifold $M$, so if we use a chart, we can get a corresponding curve $\phi\circ \gamma$ with values in the Banach space (i.e a normed vector space) $\Bbb{R}^n$, and we know how calculus works in the setting of vector spaces. So, all this map $F_{\phi,p}$ does is it takes a curve $[\gamma]$ and maps it to the "velocity vector" $(\phi\circ \gamma)'(0)$. I hope this is intuitive (otherwise, just draw a few pictures to see where each object is).
Now, it is also easy to verify that $F_{\phi,p}$ is a bijective function; I leave it to you to verify that $G_{\phi,p}:\Bbb{R}^n\to T_pM$ defined as \begin{align} G_{\phi,p}(v):= [t\mapsto \phi^{-1}(\phi(p)+tv)] \end{align} is the inverse function. In words, what we're doing is we're taking a vector $v\in\Bbb{R}^n$, and considering the straight line $t\mapsto \phi(p)+tv$. This is a curve based at the point $\phi(p)$, in the direction $v$. Since $\phi$ is a homeomorphism, it follows that for small enough values of $t$, we have $\phi(p)+tv\in \phi(U)=\text{domain}(\phi^{-1})$, hence we can consider the equivalence class of the curve $t\mapsto \phi^{-1}(\phi(p)+tv)$.
So, what has all this extra notation yielded? Well, we have a bijective function $F_{\phi,p}:T_pM\to \Bbb{R}^n$, and of course, $\Bbb{R}^n$ is a vector space, so by basic linear algebra, we can "pull back" the vector space structure of $\Bbb{R}^n$ so as to make $F_{\phi,p}$ a linear isomorphism. Explicitly, what I mean is that we can define addition and scalar multiplication $+_{\phi}$ and $\cdot_{\phi}$ (I put the subscript because everything is chart-dependent so far) as follows: \begin{align} \begin{cases} [\gamma_1]+_{\phi} [\gamma_2]&:= F_{\phi,p}^{-1}\bigg(F_{\phi,p}([\gamma_1])+ F_{\phi,p}([\gamma_2])\bigg)\\ c\cdot_{\phi}[\gamma]&:= F_{\phi,p}^{-1}\bigg(c\cdot F_{\phi,p}([\gamma])\bigg) \end{cases} \end{align}
If you unwind all the definitions, then \begin{align} c\cdot_{\phi}[\gamma_1]+_{\cdot}[\gamma_2]= [t\mapsto \phi^{-1}\left(\phi(p) + t(c\cdot (\phi\circ \gamma_1)'(0)+(\phi\circ \gamma_2)'(0))\right)] \end{align} Hopefully, the idea is clear enough: you have a bijection, so you just push everything forward, do the computations in $\Bbb{R}^n$, then bring everything back to $T_pM$, and that's how addition and scalar multiplication is defined. I leave it to you that all the vector space axioms are satisfied and that $F_{\phi,p}$ is a linear isomorphism etc.
One final thing to note is that so far the addition and scalar multiplication have been defined using a particular chart $(U,\phi)$, but actually, it is a simple chain rule exercise to verify that if you have a different chart $(V,\psi)$, then $+_{\phi}=+_{\psi}$ and $\cdot_{\phi}=\cdot_{\psi}$, so the vector space structure on $T_pM$ is actually chart-independent, hence we just denote it as $+$ and $\cdot$ as usual. I leave it to you to unwind definitions, use chain rule etc to verify this. If you have trouble, let me know, maybe I can elaborate more.