To distinguish between points and tangent vectors, let $p=(p_1,...,p_n)\in \mathbb{R}^n$ a point of $\mathbb{R}^n$ and $v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$ a point of the tangent space $\mathbb{R}^n$.
The line through $p=(p_1,...,p_n)\in \mathbb{R}^n$
with direction $v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$ has parametrization $a(t)=(p_1+tv_1,...,p_n+tv_n)$.
If f is $C^\infty$ in a neighborhood of $p\in \mathbb{R}^n$ and $v$ is a tangent vector at $p$, define the directional derivative of $f$ in the direction of $v$ at $p$ as
$$D_vf=\lim\limits_{t \to 0} \frac{f(a(t))-f(p)}{t}.$$
By the multi-variable chain rule, we have $$D_vf=\sum_{i=1}^{n} \frac{da^i}{dt}(0)\frac{\partial f}{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial f}{\partial x^i}(p).$$
Of course in the above notation $D_vf$, the partial derivatives are evaluated at $p$, since v is a vector at $p$. Now, we can define a map $D_v$ (which assigns to every
f which is $C^\infty$ the real number $D_v(f)$ ) with the natural way
$$D_v=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}\Bigr\rvert_{p}.$$
This map $D_v\in \mathcal{D}_p(\mathbb{R}^n)$ is in fact a derivation at $p$.
Finally, you can show that the map
\begin{align}
\phi :T_p(\mathbb{R}^n) &\to \mathcal{D}_p(\mathbb{R}^n) \\
v &\to D_v
\end{align}
is a linear isomorphism of vector spaces (for surjectivity, you can use Taylor's theorem).
So the answers to your questions are:
3) Υes, you can see every tangent vector $v \in T_p(\mathbb{R}^n)$ as a derivation $D_v\in \mathcal{D}_p(\mathbb{R}^n)$ using the isomorphism $\phi$.
2) Since $e_1,...,e_n$ is the canonical basis of $T_p(\mathbb{R}^n)$ and $\phi$ is an isomorphism, then $\phi(e_1),...,\phi(e_n)$ is a basis of $D_v\in \mathcal{D}_p(\mathbb{R}^n)$. But $\phi(e_i)=\frac{\partial }{\partial x^i}\Bigr\rvert_{p}$, hence $\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$
is a basis of the tangent space
$\mathcal{D}_p(\mathbb{R}^n)\simeq T_p(\mathbb{R}^n)$.
1) As a result, you can say that the basis of $T_p(\mathbb{R}^n)$ is $\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$. You can say all that because of $\phi$.
Yes, very good.
I only have three comments to make.
1) the coordinate-free definition of $\alpha \in \Omega^1(T^*M)$ is given by $\alpha_{(x,\mu)}(Z_{(x,\mu)}) = \mu({\rm d}\pi_{(x,\mu)}(Z_{(x,\mu)}))$, where $(x,\mu) \in T^*M$, $Z_{(x,\mu)} \in T_{(x,\mu)}(T^*M)$ and $\pi\colon T^*M \to M$ is the projection.
2) $\alpha$ is called the tautological $1$-form because it is the unique $1$-form in $T^*M$ such that for all $\sigma \in \Omega^1(M)$ we have $\sigma^*\alpha = \sigma$. To understand the pull-back, actually think of $\sigma$ as a section $M \to T^*M$.
3) if $\iota\colon M_0 \to T^*M$ denotes the inclusion of the zero section, we have that $\iota^*\alpha = 0$. But pull-backs commute with ${\rm d}$, so applying $-{\rm d}$ it follows that $\iota^*\omega = 0$, as you wanted.
Best Answer
Let $M$ be a smooth $n$-manifold and $(U,\phi)$ be a coordinate chart where $\phi$ has component functions $(x^i)$. Recall that $\phi$ is a smooth function $\phi:U\to\mathbb{R}^n$ whose component functions are smooth maps $x^i:U\to\mathbb{R}$. In particular, for any point $p\in M$, we may look at the differential $dx^i_p:T_pM\to T_{x^i(p)}\mathbb{R}\cong\mathbb{R}$. So for any tangent vector $v\in T_pM$, its image under $dx^i_p$ is indeed a real number.
But also recall that $dx^i_p(v)\in T_{x^i(p)}\mathbb{R}^n$ was defined to be the tangent vector such that, for any smooth $f:\mathbb{R}\to\mathbb{R}$, we have the equality $$ dx^i_p(v)f=v(f\circ x^i). $$ Thus, these equivalences of $dx^i$ essentially boil down to the fact that the tangent spaces of the smooth manifold $\mathbb{R}$ are isomorphic to $\mathbb{R}$.