Motivation for defining a tangent vector as an equivalence class of curves One definition for a tangent vector on a manifold goes like this:

We have a differentiable manifold $X$, a point $x \in X$, and two curves $\alpha, \beta:(-1,1) \to X$. Then $\alpha$ is equivalent to $\beta$ at $x$ iff $\alpha(0)=\beta(0)=x$, and for a chart $\phi:X \to \mathbb{R}^n$ we have $(\phi \circ \alpha)'(0)=(\phi \circ \beta)'(0)$. The equivalence class of all such curves is called a tangent vector.

I've got two questions, both of which may be nonsensical, but here goes.
The key piece of information here seems to be $(\phi \circ \alpha)'(0)$ (for any $\alpha$ in our equivalence class). Why bother about the curves at all when this is all we're interested in? Why not define the tangent vector to be $(\phi \circ \alpha)'(0)$, in some sense? As a linear map ($\Bbb{R} \to \Bbb{R}^n$) it does not belong to the manifold, of course, but neither does an equivalence class of curves, right?
The second question is, isn't it true that for any $v \in \Bbb{R}^n$, $v$ will be the differential of some curve(s) through $x$ (or I guess, more precisely, the value of that differential at 1)? In that case, why isn't the tangent space simply all of $\Bbb{R}^n$?
I have only the most tenuous grasp on all of this, so the simpler (and more elementary) the answer, the more likely I am to understand it.
 A: A few preliminary remarks about the definition: there is actually no need for $\alpha$ to be defined on $(-1,1)$; all you need is for $\alpha$ to be defined on a small interval $(-\varepsilon, \varepsilon)$ containing the origin $0$. Second, a chart $\phi$ may not necessarily be defined on the entire manifold $X$, only some open set $U$.

First Question:
One recurring idea in basic manifold theory is that given any familiar concepts in $\Bbb{R}^n$, we want to generalize it to the context of manifolds in a well-defined manner. Here I'm using the term "well-defined" to mean that the particular concept being developed doesn't rely on any "additional choices". In particular, we do not want the concepts of manifold theory to depend on the particular chart $(U, \phi)$ we choose.
Now, I'll address your first question. It is a bad idea to define a tangent vector at a point $x \in X$ to be an element $(\phi \circ \alpha)'(0) \in \Bbb{R}^n$, because in making this definition, you have made a choice of chart $(U,\phi)$. What if someone else comes along and they use a different chart, say $(V,\psi)$ ? Then their definition of tangent vector would be the element $(\psi \circ \alpha)'(0) \in \Bbb{R}^n$. But then we have a conflicting notion: we're talking about the same curve $\alpha$, but according to you, the tangent vector is $(\phi \circ \alpha)'(0)$, but to someone else, the tangent vector is $(\psi \circ \alpha)'(0)$... so which should it be?
I hope the above remark illustrates to you that it is the actual curve $\alpha$ which contains the key geometric information of "tangent-ness". The chart $(U,\phi)$ only comes in to allow us to make standard mundane calculations using numbers.  Therefore, from a purely logical perspective, it is a bad idea to define a tangent vector to be $(\phi \circ \alpha)'(0)$ (we run into inconsistencies).
On the other hand, it is easy to verify using the chain rule that if $\alpha, \beta : (-\varepsilon, \varepsilon) \to X$ are smooth curves with $\alpha(0) = \beta(0) = x$, then for any pair of charts $(U, \phi)$ and $(V, \psi)$ of the manifold $X$ we have
\begin{align}
\text{$(\phi \circ \alpha)'(0) = (\phi \circ \beta)'(0)$ if and only if $(\psi \circ \alpha)'(0) = (\psi \circ \beta)'(0)$}
\end{align}
In other words, if two curves $\alpha$ and $\beta$ are equivalent with respect to one chart, then they are equivalent with respect to any other chart. Therefore, the actual equivalence class $[\alpha]$ is an object which is completely-independent of charts.
I hope you realize the logic: we initially use a chart to define the equivalence relation, but after that we actually prove that the equivalence relation actually doesn't depend on the chart. Therefore the equivalence classes are actually chart-independent and  therefore qualify (from a logical perspective) as something which can be called a tangent vector at $x$.

Second Question
I hope you are now convinced from my above answer as to why a tangent vector should really only be an equivalence class of curves $[\alpha]$, rather than a vector $(\phi \circ \alpha)'(0)$. Now, you have made a very good observation, which is that every vector $v \in \Bbb{R}^n$ is the differential of some smooth curve in the manifold (technically you need to compose the curve with a chart map, and evaluate the differential at $1\in \Bbb{R}$). Let me make your informal statement more explicit:

Proposition:
Let $X$ be an $n$-dimensional smooth manifold, and fix a point $x \in X$. Suppose we are given a chart $(U,\phi)$ containing the point $x$. Now, define the function $G_{\phi,x}: T_xX \to \Bbb{R}^n$ by the rule
\begin{align}
G_{\phi,x} \big( [\alpha] \big) &:= (\phi \circ \alpha)'(0)
\end{align}
Then, $G_{\phi,x}$ is a bijective map (equivalently invertible), and its inverse $G_{\phi,x}^{-1} : \Bbb{R}^n \to T_xX$ is given by the rule
\begin{align}
G_{\phi,x}^{-1}(v) = \left[ t \mapsto \phi^{-1}(\phi(x) + tv) \right]
\end{align}

I'll leave it to you to prove the proposition, but let me highlight some (perhaps subtle) things. First of all, the function $G_{\phi,x}$ is well-defined in the sense that although we used a particular representative curve $\alpha$ belonging to the equivalence class $[\alpha]$, the RHS of $G_{\phi,x}\big([\alpha] \big) = (\phi \circ \alpha)'(0)$ doesn't depend on this choice of curve (precisely because of how we defined the equivalence relation).
Next, the expression $\left[ t \mapsto \phi^{-1}(\phi(x) + tv) \right]$ deserves some explanation. This notation means to first consider the function $\alpha$ defined by  $\alpha(t) := \phi^{-1}(\phi(x) + tv)$ (since $\phi$ is a homeomorphism, the domain of $\alpha$ contains a small neighbourhood of $0$, and you can check that $\alpha(0) = x$ and that $\alpha$ is infact a $C^{\infty}$ curve). So, the notation just means to consider the equivalence class of curves containing $\alpha$.
I'll now leave it to you check that the two functions of the proposition are actually inverses of each other. Next, you ask:

In that case, why isn't the tangent space simply all of $\Bbb{R}^n$?

Well, the short/uninspiring answer is that the elements of the tangent space $T_xX$ and $\Bbb{R}^n$ are completely different objects, therefore they cannot be equal. However, what the above proposition shows is that by CHOOSING a particular chart $(U,\phi)$ contianing the point $x$, we have established a bijection between $T_xX$ and $\Bbb{R}^n$. This is a very useful thing to know, because hopefully you recall the following basic linear algebra result (forgive my notation... I'm running out of letters):

Proposition:
Let $V$ be any set, let $W$ be a vector space over $\Bbb{R}$, and suppose we have a bijection $G:V \to W$. Then, we can define addition and scalar multiplication on $V$ such that $G$ is an isomorphism of vector spaces. More explicitly, for any $\xi_1, \xi_2 \in V$ define their sum to be
\begin{align}
\xi_1 + \xi_2 &:= G^{-1} \big(G(\xi_1) + G(\xi_2) \big)
\end{align}
and for any scalar $c \in \Bbb{R}$, and any $\xi\in V$, define
\begin{align}
c \cdot \xi := G^{-1}\big(c \cdot G(\xi) \big)
\end{align}

Therefore, using a particular chart $(U,\phi)$, we can establish a vector space structure on $T_xX$ such that the above defined map $G_{\phi,x}: T_xX \to \Bbb{R}^n$ is a linear-isomorphism. Now, we play the same game once again: in defining this vector space structure, we used a particular chart $(U,\phi)$; it is a good exercise to check that the vector space structure is actually independent of which chart is used.
Long story short, we have now made $T_xX$ into a real vector space (in a completely well-defined chart-independent manner), such that it is isomorphic to $\Bbb{R}^n$. As a nice corollary, this shows that $T_xX$ is an $n$-dimensional vector space; i.e the manifold's dimension is equal to the tangent space dimension (both are $n$). So, no $T_xX$ is not equal to $\Bbb{R}^n$, but it is isomorphic to $\Bbb{R}^n$, and if you give me a chart $(U, \phi)$, I can explicitly write down this isomorphism. So, it is in this sense that the tangent space is "almost" like $\Bbb{R}^n$.
