Can I understand $1$-form this way? It's said a function $f$ on $M$ is a $0$-form; I think an example of one form is $df=\frac{\partial f}{\partial x_1}dx^1+\dots+\frac{\partial f}{\partial dx^n}dx^n$, where $(x,U)$ is the local coordinate system, $dx_i$ is basis for tangent space $M_p$ of a manifold $M$ at p, $dx^i$ is dual basis for dual space ${M_p}^*$. (There may be something that goes wrong here, because when I read the book I am often not sure if the author is refering to vector/basis or dual vector/basis.)
Since $dx^i$ is first order tensor, so it's function of a vector, and therefore so is $df$. When $df$ acts on one-cube or other 1-dim object, say a vector $v$, we get $df(v)=df=\frac{\partial f}{\partial x_1}dx^1(v)+\dots+\frac{\partial f}{\partial dx^n}dx^n(v)$ which is a number (or  it is a number only when $f$ maps $M$ to $\mathbb{R}$ or $\mathbb{C}$ and so each partical derivative is a number?). And so $df$ is actually a one-tensor, and a measure function giving a 'length' of a vector.
We can further simplify the example, for example, let $f$ be the indentity map of $M$, then $df$ is also identity map of the tangent bundle $TM$. Then $df(v)=v$, which means $df$ is not 1-tensor or measure function. Seems to contradict my  thought above. (But if we let $f$ be a function mapping points on $M$ to numbers, then $df(v)$ gives a number. So does $0$-form $f$ have to be such?)
Then what's 1-form? Would anyone give an illuminating example?
(Since $k$-form is related to wedge product, the topic discussed here is related to my another post Is wedge product of n-dimension the length/area/volume,... of an oriented set of n vectors?, which contains a bit of discussion as well.)

It is said $k$-form is alternating covariant tensor field (alternating covariant vector field when k=1), which means it is a section of a subset $\Omega^k(TM)$ of k-fold covariant tensor field $\mathcal{T}^k(TM)$. $\mathcal{T}^k(TM)$ is a collection of functions $T:TM\times\dots\times TM\rightarrow \mathbb{R}$. So k-form should be at each point we have products of k 1-tensor, i.e. {$v_{1,p}*\otimes...\otimes v_{k,p}*, p\in M,$ and $v_{i,p}\in M_p$, i.e. is a tangent vector at $p$}. So perhaps we can roughly think that given k vector fields, changing every tangent vector to its dual vector, then we get a k-form. And it will act on k tangent vectors at one point, e.g. {$dx_{i,p}$, $1\leq i\leq k$ }, a subset of the basis of $(x,U)$ where U is a neighborhood of $p$.
Now it seems safe to say $df$ as one tensor can't be identity map of the tangent bundle $TM$. But still, how should we understand 1-form?

(Edited to add:) So 1-form $\omega$ (if exact, =$d\eta$, where $\eta:M\rightarrow \mathbb{R}^1$) is almost completely an analog of $df$ for $f:\mathbb{R}^n\rightarrow :\mathbb{R}^1$. (It and its integration are more general than a measure function of length, volume, etc.)
It's related to tensor because tensor is a linear function (to one dimensional space, whatever the order of tensor!) of several tangent vector (an analog of $dx$). So a 1-tensor is linear operator on vector(s), i.e. a $1\times n$ matrix (a transposed vector). It's therefore natural to think of k-tensor as a candidate of first order (linear approximation of) change of a function mapping vector(s) to $\mathbb{R}$.
Considering its variables,

*

*1-form is function of point of $M$ (so is $df$), an analog to the f'(x) part (when n=1) or linear operator (matrix) $A$ part of $df$, both parts being function of $x$ in n-dim Euclidean space domain.

*1-form is also a (linear) function of tangent vector, WHEN the point variable is fixed, an analog to '$\dot \ dx$' (dx as scalar or vector) part of $df$, indicating $df$ being linear transformation (i.e. matrix or having a matrix representation) of or proportional to $dx$; however, this fact looks so trivial in the usual calculus that I ignore it.

Combining this two facts, we can say 1-form is like a  (variable) matrix (probably named Jacobian matrix) ACTING on 1 vector, whose entries vary with the point variable.
In other words, we can say it is, in the usual calculus, similar to case of differential forms on manifold (1-forms is the derivative of a function between two manifolds (one is $\mathbb{R}^1$), and so maps a tangent vector to a tangent vector $\in T\mathbb{R}^1$), more comprehensive to understand differential of $f$ as $df$, rather than $f'(x)$ (when n=1) or as matrix $A$ alone (which easily causes us to think $dx$ is not part of but outside the derivative, as we do in writing $\int f'dx$ and 1-order part of function expansion $f'(x)(x'-x)$, and find the two are the same linear approx of difference.). And we can show the relation better by writing $df$ as $df_x(dx)$ where x and dx are completely independent (except that dx is at x), $df_x$ (x fixed) being a linear transformation ($t\rightarrow kt, t\rightarrow At$).
Comparing $df_x(dx)$ with the notation in diff geom, $\omega_{*p}(v)$ or $\omega_*(v_p)$, we see  the latter can really be understood as at first a function of p, and then when p is fixed, a function of (tangent) vector, i.e. of a linear approximation of displacement of point in $M$; or reversely we can understand $df$ directly as a function a $dx_x$.
Another change caused by such an switch from $df$ to 1-form perspective is that now we regard integral as (the limit of) sum of value of 1-form $\omega_{p_r}(v)$ or $df_{x_r}(dx)$ at several $p_r$ or $x_r$, that is, the linear or 1-st order approximation of change in  $\eta$ (the function whose derivative in manifold sense is $\omega$, if $\omega$ is exact) or $f$, instead of regarding integral as sum of area; since we can more conveniently obtain it given small displacement at $\omega_{p_r}(v)$ or $df_{x_r}(dx)$. Note this perspective is different from the fundamental law of calculus as it still involves sum of infinite segmentation.
We can also see $\eta$ and $f$ as scalar field on a manifold or Euclidean space.
With the new integration perspective and scalar field perspective, now we can look into, for example, length integration $L_\gamma$ of a curve in $\mathbb{R}^3$. We need to place the curve on a surface $M$ (to illustrate the vector basis' and dual basis' role; note that if we set the basis to be of 2-dim space, we have to do so). Then

*

*we can find the 1-form $\omega: \gamma'dt \mapsto \sqrt{\langle \gamma', \gamma'\rangle}dt$, i.e. $d\gamma \mapsto \sqrt{\langle d\gamma, d\gamma\rangle},\ TM\rightarrow T\mathbb{R}$, which is the derivative of $\eta:\gamma\mapsto {L_{\gamma(\tau)}}|_0^t,\ M\rightarrow \mathbb{R}$. (The expression of $\eta$ as function of an element $\in M-\gamma$ remain undefined.)

*let $\gamma=\{(x, f(x)), f:\mathbb{R}^2\rightarrow\mathbb{R}\}$, we can find the 1-form $df: ({x^1}'(t), {x^2}'(t))dt\mapsto \frac{df}{dt}dt$, i.e. $dx\mapsto df(x), T\mathbb{R}^2\rightarrow T\mathbb{R}$,
where
$$\frac{df}{dt}dt=\frac{\partial f}{\partial {x^1}}{x^1}'(t)dt+\frac{\partial f}{\partial {x^2}}{x^2}'(t)dt=\frac{df}{dt}dt=\frac{\partial f}{\partial {x^1}}d{x^1}+\frac{\partial f}{\partial {x^2}}d{x^2}.$$
We can find another one form $dg: dx\mapsto \sqrt{(df(x))^2+(dx)^2}, T\mathbb{R}^2\rightarrow T\mathbb{R}$, (here the square is norm's square).
which is the derivative of $g: x\mapsto \int_0^t \sqrt{(df(x(\tau)))^2+(dx(\tau))^2}, \ \mathbb{R}^2\rightarrow\mathbb{R}$,
where
$$\int_0^t \sqrt{(df(x(\tau)))^2+(dx(\tau))^2}=\int_0^t \sqrt{(\frac{\partial f}{\partial {x^1}}{x^1}'(\tau)d\tau+\frac{\partial f}{\partial {x^2}}{x^2}'(\tau)d\tau)^2+({x^1}'(\tau), {x^2}'(\tau))d\tau)^2}={L_{\gamma(\tau)}}|_0^t.$$ (The expression of $g$ as function of an element $\in \mathbb{R}^2-\{x|\ (x,f(x))\in \gamma\}$ remain undefined.)
So we have two 'measure' functions $\eta, \ g$ for curve length with different domain, whose derivatives are 1-forms.


New question: It seems there is an error: though $dg$ is 'integrable', it seems not to be a linear function of the vector ($dx^1, dx^2$), for the sum under $\sqrt{}$ has $\Delta=-4(f_{x_1}^2+f_{x_2}^2+1)\neq 0$. This state of 'being not 1-form but still integrable' seems weird. Is that common?
 A: You have a bunch of typos. If $f:M \to \Bbb{R}$ is smooth and $(U,x)$ is a chart on the manifold $M$ then
\begin{align}
df &= \sum_{i=1}^n \dfrac{\partial f}{\partial x^i} dx^i = \dfrac{\partial f}{\partial x^1}dx^1 + \dots + \dfrac{\partial f}{\partial x^n}dx^n
\end{align}
When we say $df$ is a $1$-form, what we mean is that for every $p \in M$, $df(p):T_pM \to \Bbb{R}$ is a linear map; i.e $df(p) \in (T_pM)^* \equiv T_p^*M$. We can also call $df$ a tensor field of type $(0,1)$. As a mere notation thing, we may sometimes write $df_p$ and sometimes $df(p)$, depending on whichever makes things easier to read.
Just so it is absolutely clear, $df$ has to be evaluated on two things before giving you a number. First you need a point $p\in M$, next you need a vector $\xi \in T_pM$. Then it is $df_p[\xi]$ (or $df(p)[\xi]$ if you wish) which is a real number.

You write

$dx_i$ is basis for tangent space $M_p$ of a manifold $M$ at p, $dx^i$ is dual basis for dual space ${M_p}^*$

We typically always write $x$ with upstairs index $x^i$, and $dx_i$ doesn't really make sense. It is $\left\{\frac{\partial}{\partial x^i}(p)\right\}_{i=1}^n$ which is a basis for the tangent space $T_pM$ (or $M_p$ as you write it) and $\{dx^i(p)\}_{i=1}^n$ which is a basis for the cotangent space $T_p^*M$.
Next, you write:

And so df is actually a one-tensor, and a measure function giving a 'length' of a vector.

No, $df$ does not measure the "length" of a vector in anyway. Taking lengths of a vector requires a norm on a vector space, or if we specialize slightly, an inner product on a vector space. On the manifold level, this is what a Riemannian metric tensor field $g$ does (it is a smooth $(0,2)$ tensor field on $M$ which is symmetric and pointwise positive-definite; i.e it is a smooth assignment $p\mapsto g(p)$ of an inner product to each tangent space $T_pM$).
What $df$ tells you is a linear approximation to the actual change in $f$ when you move in a certain direction. For example, let $\gamma: \Bbb{R} \to M$ be a smooth curve with $\gamma(0) = p$, and $\dot{\gamma}(0) = \xi \in T_pM$ (note that here $\dot{\gamma}(0)$ is obviously not the limit of a difference quotient, rather I just mean it is the tangent mapping/push-forward applied to the unit tangent vector $1_0\in T_0\Bbb{R} \cong \Bbb{R}$; in symbols $\dot{\gamma}(0):= T\gamma_0[1_0] \equiv \gamma_{*,0}[1_0]$, where the $\equiv$ means "same thing different notation"). Then, $df_p[\xi] = (f\circ\gamma)'(0)$ (this is either a definition or a very simple theorem depending on how you defined things).
So,
\begin{align}
(f\circ \gamma)(t) &= (f\circ \gamma)(0) + t (f\circ \gamma)'(0) + o(t) \tag{as $t\to 0$} \\
&= f(p) + t\cdot  df_p[\xi] + o(t)
\end{align}
So really, the $1$-form $df$ for a smooth function $f:M \to \Bbb{R}$ has the same interpretation as $Df$ (the Frechet derivative/ Jacobian matrix) does in standard multivariable calculus for a function $f: \Bbb{R}^n \to \Bbb{R}$. These are both "functions of two arguments", a base point $p$ and a vector $\xi$, and their job is to tell you the linear approximation to the actual change (which is what differential calculus is mainly about).
Therefore, if you want to properly understand $1$-forms, you need to revisit the basic multivariable calculus and the definition of derivative there and understand what it means; it's only once you understand what's happening in $\Bbb{R}^n$ (or a vector space $V$ more generally) that you can appreciate what's happening in the manifold situation.
Finally, you ask about what happens if $f:M \to M$ is the identity map. Well, in the case that the target space is not $\Bbb{R}$, I personally dislike using the notation $df$ when referring to the push-forward/tangent mapping. In general if $f:M \to N$ is a smooth map between manifolds, then I rather use the notation $Tf:TM \to TN$ to denote the tangent mapping (which using the equivalence-class of curves definition of tangent spaces sends $[\gamma] \mapsto [f\circ \gamma]$). And in the case where $f:M \to M$ is $f = \text{id}_M$, it is certainly true that $Tf = \text{id}_{TM}$. The use of $T$ makes it very memorable: $T(\text{id}_M) = \text{id}_{TM}$.
