Difficulties understanding one-forms I am currently fighting my way through The Road to reality by Roger Penrose and I have difficulties understanding one-forms. The chapter I'm currently reading is about surfaces (2-manifolds) and he introduces a function $\Phi$ locally defined on the manifold by a function $f(x,y)$ and in another coordinate patch by a different function $F(X,Y)$. 
What exactly does the quantity $d\Phi=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$ represent? Penrose says it's a derivative independent of the coordinate choice. Later he writes the one-form as a "disembodied operator form" $d=dx\frac{\partial}{\partial x}+dy\frac{\partial}{\partial y}$. I don't really understand what is meant by that and why that should represent a vector field. A vector field of what? On what do the derivative operators act and what does $dx$ and $dy$ mean? I'd be glad if someone could explain to me the basic notion behind all of this.
 A: I assume you don't want to see all the rigorous mathematics stuff. I will explain this in a plain context. I will use Einstein summation as I assume you are familiar with theoretical physics.
The manifold is only local homeomorph to $\mathbb R^n$, so depending on the choice of basis vector, you will have some corresponding coordinates $(x^1,x^2,...,x^n)$.These coordinates has a huge problem: They are not unique. i.e. different points can have same coordinate in different chart.
To fix that, one introduce the coordinate system. A coordinate system is simply a set of four differentiable scalar fields $x^{\mu}_X$ that attach a unique set of labels to each point X. 
Now each point on the manifold can be labeled uniquely by a set of numbers $(x^1,...,x^n)$ we are provided with a special set of basis vectors which fulfill the requirement:
Suppose we have $x^{\mu}+dx^{\mu}$. The infinitesimal difference vector between the two points, denoted $d \bar x$ is a vector defined at the point $X$. We define the coordinate basis as the set of four basis vectors $e_{\mu X}$ such that the components of $d\bar x$ are $dx^{\mu}$:
$$d\bar x:=dx^{\mu}e_{\mu}$$
You can think $x^{\mu}$ as kind of global coordinates based on a collection of charts. (Just as you are familiar in a flat space $(x,y)$, only here you will have additional problem with connections) Therefore, if you are differentiating, you are differentiating in these coordinates.
Consider
$$df=\frac{\partial f}{\partial x^{\mu}}dx^{\mu}$$ will be a scalar independent of coordinate choice.(Treating f as a function of the coordinates) As I mentioned before, the $dx^{\mu}$ are components of vector. Then the gradient should be element from one-form.
The differential that you are talking about is the gradient. It is kind of linear function of tangent vector($dx^{\mu}$). The operator is just a mapping. Mathematically speaking the operator $d$ defines a mapping $$d:C^{\infty} \rightarrow T_M^*$$ where $T_M^*$ is the cotagent space. This means, it needs a vector field to act on. $df$ can be then written as $<\nabla f,d\bar x>$, where $d\bar x$ is an element from tangent space.
