Why such cotangent space have $dx_i$ instead of $d/dx_i$ as part of their basis? 
DEFINTION OF TANGENT BUNDLE:
Given a smooth manifold $M \subset \mathbb{R}^{n}$ embedded as a
hypersurface represented by the vanishing locus of a function $f \in
 C^{\infty}\left(\mathbb{R}^{n}\right)$ with the condition that $\nabla
 f \neq 0$, the tangent bundle is
$$ T M=\left\{(x, v) \in T \mathbb{R}^{n}: f(x)=0, d
f_{x}(v)=0\right\} $$

My question is since $d f_{x}(v)=\nabla f(x) \cdot v=0$ so doesnt this mean $v=(dx_1,...,dx_{n-1})$ assuming a point in $M$ has the local coordinate representation $(x_1,...,x_{n-1})$. I am writing $v=(dx_1,...,dx_{n-1})$ under the impression that if I expand $\nabla f(x) \cdot v$ I get the form $\sum df/dx_i \cdot dx_i$ and so I have grouped up $dx_i$ to form $v$, since as we know, gradient is perpendicular to the hypersurface so $v$ must be the tangent.
$ \Rightarrow$ $TM$ has the representation $(x_1,...,x_{n-1},dx_1,...,dx_{n-1})$
If this is the case, why do we write the 1-forms which belongs to the space of $T^*M$ as $$\alpha=\sum_{i=1}^{n-1} \xi_{i} d x_{i}$$ where $\xi_i \in \mathbb{R}$. Shouldn't it be something like $$\alpha=\sum_{i=1}^{n-1} \xi_{i} \frac{d}{d x_{i}}$$
Where did I go wrong?

The idea behind me asking this question is, I understand the directional derivative definition of tangent vectors (I might be wrong here), but as soon as I view it from the perspective of my manifold being represented by $f(x)=0$ on some bigger embedding then directional derivative seems to take up a different meaning.

Alternatively,
How does one derive that a 1-form can be expressed as  $$\alpha=\sum_{i=1}^{n-1} \xi_{i} d x_{i}$$
I should add that I am not a vigorous mathematician, so its very likely that I might be defining/interpreting something wrong.
 A: $\newcommand{\dd}{\partial}$There's an important distinction between "ambient" and "intrinsic" coordinates. This may not exactly answer your question, but is too long for a comment.
Let's look at the unit circle $M$ as an example:
$$
f(x_{1}, x_{2}) = x_{1}^{2} + x_{2}^{2} - 1 = 0.
$$
Neither ambient coordinate $x_{1}$ or $x_{2}$ is a global coordinate, but at each point of $M$, either $x_{1}$ or $x_{2}$ (or at most points, both) can be used as a local intrinsic coordinate.
Differentiating the preceding equation gives
$$
df(x_{1}, x_{2}) = 2x_{1}\, dx_{1} + 2x_{2}\, dx_{2} = 0.
$$
If $(x_{1}, x_{2})$ is a point of the circle, then a vector $v = (v_{1}, v_{2})$ in the plane is tangent to $M$ at $(x_{1}, x_{2})$ if and only if
$$
df(x_{1}, x_{2}) v = 2x_{1}\, dx_{1}(v) + 2x_{2}\, dx_{2}(v)
= 2x_{1}v_{1} + 2x_{2}v_{2} = 0.
$$
The components of $v$ are not $dx_{1}$ and $dx_{2}$ themselves, but the values of these ambient coordinate $1$-forms on $v$. In this example, the vector $v$ is tangent to $M$ at $(x_{1}, x_{2})$ if and only if there is a real number $\lambda$ such that $(v_{1}, v_{2}) = \lambda(-x_{2}, x_{1})$.
What about coordinate vector fields? If we use $x_{1}$ as local coordinate (which we're free to do except at the points $(\pm1, 0)$ where the tangent line is vertical), then in this example $x_{2} = \sqrt{1 - x_{1}^{2}}$ on the upper semi-circle, or $x_{2} = -\sqrt{1 - x_{1}^{2}}$ on the lower semi-circle.
Since $\frac{\dd}{\dd x_{1}}$ has horizontal component $1$, we have $\frac{\dd}{\dd x_{1}} = (1, \mp\frac{x_{1}}{\sqrt{1 - x_{1}^{2}}})$, with the sign chosen depending on location. Every vector field on the circle may be written in an $x_{1}$-coordinate neighborhood as $v_{1}(x_{1})\, \frac{\dd}{\dd x_{1}}$ for some function $v_{1}$, or as $(x_{1}, v_{1}(x_{1}))$. (Note: There is only one summand because the circle is $1$-dimensional, even though there are two ambient coordinates.)
Expressing vector fields in terms of the local coordinate $x_{2}$ (which we may do except at $(0, \pm1)$) is similar.
These manifold coordinates obtained by restricting ambient coordinates are not especially elegant, but the point is to illustrate how coordinate notation works in this situation.
A: It's really all just choices of notation, so you have to be very explicit with your notation. However, you have a totally valid confusion here, and this is a common issue people have with the gradient.
Technically $\nabla f$ is a vector field, not a $1$-form, so it doesn't actually make sense to apply $\nabla f$ to a tangent vector. However, we can still take the dot product $\nabla f\cdot v$. It only makes sense to take dot products with vectors in the same vector space, so actually the basis vectors of $\nabla f$ and $v$ should be the same.
To make this more natural, we really want to use $df$, which is dual to $\nabla f$. In coordinates, $\nabla f$ is a column vector, while $df$ is a row vector. We can then apply $df$ to a tangent vector, and this corresponds to multiplying a $1\times n$ matrix with an $n\times 1$ matrix. So the basis of $df$ should be dual to that of $v$.
