Basis for the Tangent space and derivations at a point

$\newcommand{\R}{\mathbb{R}}$ 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 $\varphi^{-1}(v) = \sum_{i=1}^n v^i e_i$.

I think what you want follows from the definition of the "partials" in an arbitrary manifold. If I misunderstood your question, please advise, and I will delete my answer.

Let $x\in \mathbb R^n$, and $r_i$ the projections from $\mathbb R^n$ to the $i^{th}$ coordinate. As you point out, Tu proves that the derivations $\{\partial/\partial r_i\}_x$ form a basis for $T_x\mathbb R^n$.

Now let $p\in M$, an $n$-dimensional manifold and consider a chart $(\phi,U)$ about $p$. Then, since $\phi$ is a diffeomorphism onto $\phi(U),\ \phi_*^{-1}: T_{\phi(p)}R^n\to T_pU\cong T_p M$ is an isomorphism.

Now, if we $\textit{define}\ \left(\partial/\partial x_i\right)_p$ to be $\phi_*^{-1}\left(\partial/\partial r_i\right)_p,$ we get a basis for $T_pM$.

From this definition, the rest follows: the covectors $dx^i$ are simply the (dual) basis for $T_p^*M.$