Let $M$ be a smooth Manifold.
Let $p \in M$.
Let $U$ be an open neighborhood of $p$.
We denote by $C^{\infty}(U)$ the set of real valued smooth functions on $U$.
Let $\Lambda_p = \bigcup C^{\infty}(U)$, where $U$ runs through all open neighborhoods of $p$.
Let $f, g \in \Lambda_p$.
Suppose $f \in C^{\infty}(U)$ and $g \in C^{\infty}(V)$.
If there exists an open neighborhood $W$ of $p$ such that $W \subset U \cap V$ and $f|W = g|W$,
we say $f$ and $g$ are equivalent.
This is an equivalence relation on $\Lambda_p$.
We denote by $\mathcal{O}_p$ the set of equivalence classes on $\Lambda_p$.
Clearly $\mathcal{O}_p$ is an $\mathbb{R}$-algebra.
Let $f \in C^{\infty}(U)$, where $U$ is an open neighborhodd of $p$. We denote by $[f]$ the equivalence class containing $f$.
A derivation of $\mathcal{O}_p$ is a linear map $D\colon \mathcal{O}_p \rightarrow \mathbb{R}$ such that
$D(fg) = D(f)g(p) + f(p)D(g)$ for $f, g \in \mathcal{O}_p$.
The set $T_p(M)$ of derivations of $\mathcal{O}_p$ is a vector space over $\mathbb{R}$ and is called the tangent space at $p$.
Let $\epsilon > 0$ be a positive real number.
We denote by $\Gamma_p(\epsilon)$ the set of smooth curves $\gamma \colon (-\epsilon,\epsilon) \rightarrow M$ such that $\gamma(0) = p$.
Let $\Gamma_p = \bigcup_{\epsilon>0} \Gamma_p(\epsilon)$.
Let $(U, \phi)$ be a chart such that $p \in U$.
Let $\gamma_1, \gamma_2 \in \Gamma_p$.
Then $\gamma_1$ and $\gamma_2$ are called equivalent at $0$ if $(\phi\circ\gamma_1)'(0) = (\phi\circ\gamma_2)'(0)$. This definition does not depend on the choice of the chart $(U, \phi)$.
This defines an equivalence relation on $\Gamma_p(M)$.
Let $S_p(M)$ be the set of equivalence classes on $\Gamma_p(M)$.
For $\gamma \in \Gamma_p(M)$, we denote by $[\gamma]$ the equivalence class containing $\gamma$.
We will define a map $\Phi\colon S_p(M) \rightarrow T_p(M)$.
Let $c \in S_p(M)$.
Choose $\gamma \in \Gamma_p(M)$ such that $c = [\gamma]$.
Let $f \in C^{\infty}(U)$, where $U$ is an open neighborhood of $p$.
We write $D_c([f]) = (f\circ\gamma)'(0)$ for $f \in C^{\infty}(U)$.
Clearly $D_c$ is well defined and does not depend on the choice of $\gamma$.
Clearly $D_c \in T_p(M)$.
Hence we get a map $\Phi\colon S_p(M) \rightarrow T_p(M)$ such that $\Phi(c) = D_c$.
We claim that $\Phi$ is bijective.
Let $c, e \in S_p(M)$.
Suppose $D_c = D_e$.
Suppose $c = [\gamma]$ and $e = [\lambda]$.
Let $(U, \phi)$ be a chart such that $p \in U$.
Let $\pi_i:\mathbb{R}^n \rightarrow \mathbb{R}$ be the $i$-th projection map: $\pi_i(x_1,\dots,x_n) = x_i$.
We denote by $\phi^i$ by $\pi_i\circ\phi$.
Since $D_c([\phi^i]) = D_e([\phi^i])$, $(\phi^i\circ\gamma)'(0) = (\phi^i\circ\lambda)'(0)$.
Hence $(\phi\circ\gamma)'(0) = (\phi\circ\lambda)'(0)$.
Hence $\gamma$ and $\lambda$ is equivalent.
Thus $\Phi$ is injective.
Let $D \in T_p(M)$.
Let $(U, \phi)$ be a chart such that $p \in U$.
We assume that $\phi(p) = 0$.
We define $\phi^i$ for $i = 1,\dots,n$ as above.
Let $D([\phi^i]) = a_i$ for $i = 1,\dots,n$.
There exists $\epsilon > 0$ such that $(a_1t,\dots,a_nt) \in \phi(U)$ for every $t \in (-\epsilon, \epsilon)$.
Let $\gamma(t) = \phi^{-1}(a_1t,\dots,a_nt)$ for $t \in (-\epsilon, \epsilon)$.
Then it's easy to see that $\Phi([\gamma]) = D$.
Hence $\Phi$ is surjective and we are done.
Best Answer
For smooth manifolds the tangent space given by equivalence classes of smooth curves and the set of derivations on smooth functions is equivalent. You can give an isomorphism between these constructions. Each has its advantages.
For example, in the equivalence class of curves set-up the differential is really simple; $dF(\gamma) = F \circ \gamma$ the mapping $F: \mathcal{M} \rightarrow \mathcal{N}$ pushes the curve on $\gamma$ on $\mathcal{M}$ to the curve $F \circ \gamma$ on $\mathcal{N}$. To show $d(F \circ G)=dF \circ dG$ for manifolds we can argue $$ d(F \circ G)(\gamma) = F( G (\gamma)) = F( dG(\gamma)) = dF(dG(\gamma)) = (dF \circ dG)(\gamma) $$ I think the proof in the derivation view is less slick.
Derivations naturally arise from coordinates systems as the partial derivatives with respect to the manifold coordinates. The differential again pushes vectors from one manifolds tangent space to another. Or from the manifold to itself, but where the coordinate system has two overlapping charts. Believe it or not, you've already done this calculation in multivariate calculus. Changing coordinates from cartesian to polar for example: $$ \frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} +\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} = \cos(\theta)\frac{\partial}{\partial r}-\sin(\theta)\frac{\partial}{\partial \theta}$$ This can be understood as the vector $\frac{\partial}{\partial x}$ being pushed forward to the vector $\cos(\theta)\frac{\partial}{\partial r}-\sin(\theta)\frac{\partial}{\partial \theta}$ by the differential of the transition map. In other words, there are constructions for which the derivation formulation of tangent space nicely dovetails. Usually the derivation is sold as a sort of directional derivative like object. The formula $X(f) = \sum_{i=1}^n X^i \frac{\partial f}{\partial x^i}$ is like $D(f)(\vec{X}) = (\nabla f) \cdot \vec{X}$.
In either case, it is a bit unsettling to trade seemingly static objects like geometric vectors for classes of curves or differential operators. There is at least one other popular view, the contravariant vector formulation of tangent vectors.
It should be commented that when we consider $C^k$ manifolds this correspondence breaks down.See page 49 in Conlon's 2nd edition of Differentiable Manifolds. In his Lemma 2.2.20 he builds functions which are used to frame the isomorphism between curves and derivations. The construction fails for $C^k$ manifolds. Moreover, the reason for this is that the space of derivations on $C^k$-germs of functions is infinite dimensional.
I'd love to say you can pick one to understand. However, the nature of the subject of differential geometry requires you to be conversant in all these views.