If a manifold has curvature, then parallel transport of vectors depends on the path. In other words, any map between tangent spaces at different points is dependent on an arbitrary choice which is particular for each pair of points and each particular manifold. It is the same problem as when relating a finite-dimensional vector space $V$ with its dual $V^*$. In order to do so, we need to fix a basis for $V$.
In regards to the importance of naturality, informally a transformation is called natural if it is "independent of its source" in a sense. For example, the isomorphism between $V$ and $V^{**}$, the double algebraic dual, is natural. We send $v\in V$ to $v^{**}\in V^{**}$ such that $v^{**}(f)=f(v)$ for all $f\in V^*$. Note that this is independent not only of the basis of $V$, but also does not reference anything particular about $V$, such as a basis, and so this isomorphism can be carried out consistently over all finite-dimensional vector spaces over a common field. Quoting Wikipedia, a transformation is not natural if it cannot be extended consistently over the entire category in question.
1) Taking differential derivatives allows you to do differential calculus on manifolds. One explicit example could be defining tangent fields, i.e. maps $X:M\to TM:=\sqcup_{p\in M}T_pM$ such that $\pi\circ X=\mathrm{id}_M$ where $\pi:TM\to M$ is the canonical projection, and integrating them in order to get flow maps, i.e. maps $\varphi:\mathbb{R}\times M\to M$ such that $\varphi(0,\cdot)=\mathrm{id}_M$ and $\left.\frac{\partial\varphi(\cdot,x)}{\partial t}\right|_t=X_{\varphi(t,x)}$. Thus, from linear data ($X$), you recover a family of diffeomorphisms of $M$ with a certain behaviour.
2) If your manifold $S$ is a submanifold of an ambient one $M$, the inclusion $i:S\to M$ induces a map $di_p:T_pS\to T_pM$ which allows you to consider the tangent space of $S$ at $p$ as a linear subspace of the tangent space of $M$ at $p$. There is an other identification for tangent vectors of affine manifolds (that is $M=\mathbb{R}^n$ with the maximal atlas induced by $\mathcal{A}=\{(\mathrm{id}_{\mathbb{R}^n},\mathbb{R}^n)\}$) in order to identify them with actual vectors of $\mathbb{R}^n$: this identification is given by $\mathbb{R}^n\ni v\mapsto\partial_v\in T_p\mathbb{R}^n$, where $\partial_v$ acts on functions $f\in C^\infty_p(\mathbb{R}^n)$ by
$$\partial_vf=\lim\limits_{t\to 0}\frac{f(p+tv)-f(p)}{t}.$$
In other words, you identify the vector $v$ with the directional derivative in the direction $v$. So when you have a submanifold $S$ of an affine one, you can:
Identify a tangent vector of $S$ as a tangent vector of $\mathbb{R}^n$
Identify the tangent vector of $\mathbb{R}^n$ with an actual vector of $\mathbb{R}^n$.
3) Again, taking directional derivatives on a manifold is authorizing himself to do differential calculus on manifolds, allowing the use of useful theorems as implicit function theorem or inverse function theorem. For the identification of the two definitions, I will answer it in 4).
4) You answer your question by pointing the identification $[\gamma]\mapsto D_\gamma$, but you have to be carful that this does not depend of the choice of the representant $\gamma$. But since
$$(f\circ\gamma)'(0)=(f\circ\varphi^{-1}\circ\varphi\circ\gamma)'(0)=d(f\circ\varphi^{-1})_{\varphi\circ\gamma(0)}\left((\varphi\circ\gamma)'(0)\right)$$
by the chain rule, it is clear by the definition of the equivalence relation that is will be the case.
Best Answer
$\newcommand{\R}{\mathbb{R}}$ So because we are in $\mathbb{R}^n$ we can all imagine (or at least were told) how we should visualize what a tangent space is at a certain point $p$ in $\mathbb{R}^n$, i.e. some arrows that are tangent. The key point here is that this concept only holds because one can think of the $\mathbb{R}^n$ as vector space (and usually does it). Then a point in $\R^n$ is really nothing else then a vector in $\R^n$. But this is something that will not be true for manifolds. So when you think of $\R^n$ as being a manifold you should not think of any point in $\mathbb{\R}^n$ as vector but really just as point. Now, you can attach a vector space, that is the tangent space, at any point in $\mathbb{R}^n$, and then from there you have a vector space $T_p(\mathbb{\R}^n)$ associated with that point $p$ of the manifold $\R^n$.
Of course now that seems like just a lot of words and really for the $\R^n$ it probably is, but as we get more and more abstract those difference are key in the theory of manifolds.
Now, think of a donut or a sphere in $\R^n$, then again you can attach to any point in that manifold a tangent space. Intuitively you will know how to do it, because that object is imbedded into a euclidian space. But later on in your studies of manifolds this will not be the case! (At least not in any practical way).
So, we need to find a concept of tangent vectors that only rely on the local properties of a point of a manifold. And this is were the (partial) derivative notation comes into play. Because these are concepts that can be also defined for manifolds in general.
To the question at hand: As it was pointed out, isomorphic means it is the same thing! Both are $n$ dimensional linear vector space which are isomorphic to each other and now the only question is which kind of "names" you want to give these vectors. But that is all that is different: the symbols.
Edit: Maybe to make it more precise: Take the basis $e_1,\dots,e_n$ of $T_p(\R^n)$ and let $\varphi : T_p(\R^n) \to D_p(\R^n)$ be the isomorphism. Then $\varphi(e_i) = \partial/\partial_{x^i}|_p$ for all $i\in\{1,\dots,n\}$. Now, you do the calculations for this basis in $D_p(\R^n)$. So, whenever you now have a element $v \in D_p(\R^n)$ and you had enough of this derivative notation you simply do $\varphi^{-1}(v)$ or even more concrete: you have $v = \sum_{i=1}^n v^i \partial/\partial_{x^i}|_p$, so $\varphi^{-1}(v) = \sum_{i=1}^n v^i e_i$.