The Riemannian metric provides a natural identification of the cotangent space with the tangent space. Therefore the cotangent space is also naturally equipped with a Euclidean (or if you prefer Hilbert-space) metric. The simplest way of thinking of the identification is to send a tangent vector $v$ to the covector given by the inner product with $r$, namely $\langle v,\cdot\rangle$.
So they are naturally isomorphic as Hilbert spaces, Banach spaces, and metric spaces.
Expressed in terms of indices, the relation is simply this. If $v_i$ represents a vector and $\alpha^j$ a covector then the relation $v_i=g_{ij}\alpha^j$ (summation over a repeated index as usual) gives a way of passing between a vector and its corresponding covector. Here $g_{ij}$ is the Riemannian metric.
$\newcommand{\R}{\mathbb{R}}$
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 $\varphi^{-1}(v) = \sum_{i=1}^n v^i e_i$.
Best Answer
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.