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It seems 1-form on a manifold (let's limit our discussion to be a 2-manifold) is derivative $d\eta$ of a function $\eta:M\rightarrow \mathbb{R}$. It's similar to the derivative of $f:\mathbb{R}^2\rightarrow \mathbb{R}$, i.e. $df=df(x)=f'(x)dx$, which is in diff geom notation, $df_x(dx)$, a linear approximation of change of $f$'s value at point $x$ with displacement dx. (right?)

Then what's the analog of 2-form for $f:\mathbb{R}^2\rightarrow \mathbb{R}$? e.g. something like $\int\int f_xf_ydxdy+f_yf_xdydx+f_xf_xdxdx+f_yf_ydydy$ ($f_x$ etc. mean partial derivative)?

(Edited to add:)

This section is a note for myself:

Since a 2-form can be derived from 1-form $\omega$ in the following way:

$X(\omega Y)+Y(\omega X)+\omega[X,Y]$

all three items are like '1-form times or acting on a vector field', I guess 2-form would be like (especially when $\omega$ is exact), or more specifically like a linear combination of differential (in Lie derivative sense; by Lie derivative we get sth of the same kind) of

$$\sum \frac{\partial f}{\partial x^i} (\sum a^i\frac{\partial }{\partial x^j}).$$

Since 2-form is a (covariant tensor) field, its value at a point $p$ will be like

$$\sum (\frac{\partial }{\partial x^i})_p\ f \cdot (\sum (a^j)_p(\frac{\partial }{\partial x^j})_p),$$

where ${x^I}_p$ is the coordinate system at $p$.

So to understand 2-form perhaps I need to understand vector field/tangent bundle and its basis as well as Lie derivative, and do some calculation about them in a coordinate system.

Charlie Chang
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    A derivative $d\eta$ is an example of a 1-form. A 1-form in general is a field of row-vectors with respect to some basis. What you wrote is not a 2-form because it has to be linear operator acting on $f$. It could be something like $\partial_x\partial_y-\partial_y\partial_x$. As you can see this gives 0 on scalar functions $f$, but it gives non-zero results on vector fields. Again in general, a 2-form is not a derivative but a field of 2-tensors $A_{ij}$ which are anti-symmetric in the indices. – Chrystomath Jul 28 '20 at 09:12
  • By saying 'row-vectors' I guess you mean linear transformation of a vector (to a scalar), i.e. 1-tensor, it just happens to be a derivative of something sometimes (when it's exact) (similar to when a locally (i.e. $x$ fixed) linear function of displacement $dx$ can be a differential of something? for example, $udx_1+vdx_2$ is 1-form but not a derivative if $u_y\neq v_x$. Such cases exist only when the domain is of dim greater than 1). – Charlie Chang Jul 28 '20 at 09:31
  • Yes precisely. I don't know how much you know. A $k$-form that is exact can be written as $d\phi$ where $phi$ is a $(k-1)$-form. – Chrystomath Jul 28 '20 at 09:34
  • Well, I still need to look into 2-form to make my understanding of it clearer. – Charlie Chang Jul 28 '20 at 09:36

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