What is the motivation of creating of $T^*_p(\mathbb R^n)?$ How can we visualize covectors? 
Question 1
In calculus, we visualize the tangent space $T_p(\mathbb R^n)$ at $p$ in $\mathbb R^n$ as the vector space of all arrows emanating from $p$. What is the motivation of creating of $T^*_p(\mathbb R^n)?$ How can we visualize covectors? 
Question 2
How do I prove the sets $T^*_p(\mathbb R^n)$ are all disjoint? where $p\in U$
Considering $p,q\in U$ and $p\neq q$, suppose f is a non zero linear functional and $f \in T^*_p(\mathbb R^n)\cap T^*_q(\mathbb R^n)\implies f\in T^*_p(\mathbb R^n)$ and $f\in T^*_p(\mathbb R^n)$.How do I proceed further?. Can you please help me?
 A: I only given an answer to the first question. My answer is an addition to the answer given by janmarqz. Basically I just provide some plots for his second point. 
Visualization
I like the visualization given in V. Arnolds, Mathematical Methods of Classical Mechanics.
Suppose we have a covector $\omega \in T^*_0\mathbb{R}^2$, we then draw the hyperplanes $\dots, \omega^{-1}(\{-1\}),\omega^{-1}(\{0\}),\omega^{-1}(\{1\}), \omega^{-1}(\{2\}), \dots$
(Let's call them the contour lines of $\omega$.)
If we now want to see how the covector acts on a vector $\mathbf{v} \in T_0\mathbb{R}^2$, we just draw the arrow representing $\mathbf v$ and look how many contour lines we hit. 
Below I have drawn this method for a basis $\{\omega, \eta\}$ of $T^*_0\mathbb{R}^2$.

Why don't we draw arrows for covectors?
A typical difference between tangent vectors and cotangent vectors is their transformation behavior!
Let's use the linear transformation $\Phi:\mathbb R^2 \to \mathbb R^2: (x,y)^T \mapsto (x,2 y)^T$. 
Vectors transform via the differential $T\Phi: T \mathbb R^2 \to T \mathbb R^2$ (sometimes also denoted by $D\Phi$). In our case this is just the Jacobian of $\Phi$ i.e.
$\mathbf v \mapsto \begin{pmatrix}1&0\\0 & 2\end{pmatrix} \mathbf v$.
Covectors transform differently! An example is the gradient of a smooth function $f:\mathbb R^2 \to\mathbb R$. In classical calculus, we have see, that $\nabla f$ is always orthogonal to the contour-lines of $f$! But if you apply the transformation $\Phi$ and transform $\nabla f$ like a vector, it will not be orthogonal anymore!

But, if we visualise covectors by their contour-lines, they behave correctly! For example, if we draw the contour-lines of $df$, they will stay tangent to the contour-lines of $f$ after applying a transformation to the visualization, which is the correct behavior! 
This is one of the reasons, why we treat vectors and covectors differently. And also the reason why we draw contour-lines of covectors instead of arrows.
Higher Dimensions
The reasoning above is still valid in higher dimensions, here the contour-lines become hyperplanes instead. (The example with the gradient is only plausible for Riemannian manifolds, but the example should just give a feeling how covectors are different than vectors. In general covectors transform according to the inverse of the adjoint of $T\Phi$, i.e. $\omega \mapsto ((T\Phi)^*)^{-1}$. )
A: Question 2 is easy: We already consider the tangent spaces $T_p{\mathbb R}^n$ as disjoint: Each point $p$ has its own tangent space. Of course all these tangent spaces are carbon copies of one and the same "model space" ${\mathbb R}^n$. (If we would not make the spaces $T_p{\mathbb R}^n$ disjoint a vector field would become a hedgehog!) Now each $T_p{\mathbb R}^n$ has its own dual space $\bigl(T_p{\mathbb R}^n\bigr)^*$, and these dual spaces are considered to be disjoint as well. This is matter of definition; no proof is needed.
Question 1 is more difficult. Number 1 there is no universally agreed way to "draw" a cotangent vector in a picture showing the vector space $V$, e.g., $T_p{\mathbb R}^n$. Sometimes you have a scalar product in $V$. It is then possible to identify each $\phi\in V^*$ with a vector $a_\phi\in V$, such that the identity $$\phi(x)=a_\phi\cdot x\qquad\forall\>x\in V$$
holds. E.g., if $\phi:=df(p)$ is the differential of a scalar function $f$ at $p$ then $a_\phi=\nabla f(p)$.
If you want an "intuitive" description of covectors $F\in V^*$ think of  physical forces. Physicists cannot explain what a force is, for the same reasons that you have no intuitive feeling for covectors. They then draw the vector $a_F\in V$ and call it the force vector. In reality an (e.g., gravitational) force acting at $0\in V$ is a covector $F\in V^*$. If you want to move along the segment $\sigma$ connecting $0\in V$ with the point $x\in V$ then  you have to perform the work $W_\sigma=F.x$ (or get back $W_\sigma$, depending on the sign conventions), whereby the $.$ denotes evaluation.
A: There are few steps to consider:
1) How are the level sets of a linear maps $\Bbb R^2\to\Bbb R$?
2) How are the level sets of a linear maps $\Bbb R^3\to\Bbb R$?
3) How are the level sets of a linear maps $\Bbb R^n\to\Bbb R$?
4) The tangent spaces $T_p\Bbb R^n$ are vector spaces associated to a position $p$.
5) Different points in $\Bbb R^n$ receive different tangent spaces so they are disjoint.
6) The dual of a tangent space $T^*_p{\Bbb R^n}$ are linear maps $T_p{\Bbb R^n}\to\Bbb R$, so a 1-form is an association of a linear map to a copy of $\Bbb R^n$ attached to $p$ in which one gets the picture explained in points 1), 2) or 3) above. 
What about an example?
Let's illustrate with a simple one-form $\omega=(5+x^2-y)dx+(9+x+y^2)dy$. With it you can prescribe to a points in $\Bbb R^2$ a linear functional, let's pick at the origin $\omega|_{0,0}=5dx+9dy$
which is the same as the linear map in $\Bbb R^2\to\Bbb R$ given by $$(x,y)\mapsto\omega|_{0,0}(x,y)=5x+9y$$
Here we can experiment a see how a little horizontal variation, say at $(\epsilon,0)$ what covector gonna be?
This will be
$$\omega|_{\varepsilon,0}(x,y)=(5+\varepsilon^2)x+(9+\varepsilon)y$$
which is $\approx5x+9y$, for $0<\varepsilon<\!\!<1$.
