Coordinate-free definition of tangent space of differentiable manifold

I am working on some notes about differentiable manifolds and I am trying to introduce the concept of tangent space at a point $P\in\mathbb{M}$ without making explicit reference to a coordinate patch. I consider the set $C(P)$ of all (smooth as necessary) curves $\gamma: I\in\mathbb{R}\to\mathbb{M}$ such that $P\in \gamma(I)$. Since it is always possible to shift $I$ by a constant, I will assume that $\gamma(0)=P$. On such set of curves, I introduce an equivalence $\sim$ as follows: $\gamma\sim\beta$ if $\forall f:\mathbb{M}\to\mathbb{R}$ (smooth as necessary) $$ \frac{d(f\circ \gamma)}{dt}|_{t=0}=\frac{d(f\circ \beta)}{dt}|_{t=0}. $$
I define the tangent space as the quotient space $C(P)/\sim$. Next, I need to show that $C(P)/\sim$ is a vector space. The zero element is the equivalent class of the constant curve $\gamma(t)=P$. Let $\alpha$ be a real scalar, $\gamma'\in C(P)/\sim$, and $\gamma$ a curve in the equivalence class $\gamma'$. Then $\alpha\gamma'$ is the equivalence class that contains $\gamma(\alpha t)$. So far, so good. The problem I am having is how to construct the equivalence class of the sum of two equivalence classes without falling back to considering the representation of the curves in a coordinate patch. In other words, how can I define the sum in a coordinate independent way? Thank you for your time

You can't. The problem is that you consider equivalence classes of smooth curves $\gamma : (I,0) \to (M,p)$. To add such elements you must essentially associate to any two curves $\gamma_i : (I_i,0) \to (M,p)$ a sum curve $\gamma_1 + \gamma_2 : (I_3, 0) \to (M,p)$. But there is no additive structure on a general $M$, it only works if $M$ is an open subset of some normed vector space and $p = 0$. This is where charts enter and you cannot avoid it. Although $\gamma_1 + \gamma_2$ depends on the choice of a chart, the equivalence class of $\gamma_1 + \gamma_2$ does not and it turns out that $[\gamma_1] + [\gamma_2] = [\gamma_1 + \gamma_2]$ is well-defined.

In contrast scalar multiplication works as you define it. The reason is that scalar multiplication only involves a single direction and you can shift multiplication with $\alpha \in \mathbb R$ to the domain $I$ which has a linear structure, thereby changing the speed of $\gamma$.

I do not think that this is a real problem. Of course it would be nice to define the sum in a coordinate independent way, but recall that the concept of a smooth map itself cannot be defined in a coordinate independent way.

There are other approaches to introduce the tangent space which automatically give you the structure of a vector space. As Didier comments, you can define tangent vectors as derivations on the algebra of germs of real-valued smooth functions at $p \in M$, or alternatively as derivations on the algebra of real-valued smooth functions on $M$. Since the range $\mathbb R$ of these functions is a vector space, you get a natural vector space structure on the set of derivations.

Anyway, even if you have obtained the desired vector space structure on $T_pM$, it is not a priori clear what its dimension is. Theoretically $T_pM$ could be infinite-dimensional. And here, again, we need charts to prove that $\dim T_pM = \dim M$. This step is much easier if you work with equivalence classes of smooth curves.

Perhaps you also find it useful to have a look at Different definitions of the tangent space at a point of a smooth manifold: Biduals?