In A Geometric Approach to Differential Forms by David Bachman you can find the passage:
... a 1-form is a linear function which acts on vectors and returns numbers. For the moment let's just look at 1-forms on $T_p\mathbb R^2$ for some fixed point, $p.$ Recall that a linear function $\omega,$ is just one whose graph is a plane through the origin. Hence, we want to write down an equation of a plane through the origin in $T_p\mathbb R^2 \times \mathbb R,$ where one axis is labelled $dx,$ another $dy,$ and the third $\omega.$ This is easy: $\omega=a\; dx + b \;dy.$ Hence, to specify a 1-form on $T_p\mathbb R^2$ we only need to know two numbers: $a$ and $b.$
I am aiming at an intuitive idea of what he is referring to. So far this would be my summary:
1-forms are elements of $V^*$ that "eat" a vector to produce a scalar. In general I picture them as row vectors, but in the particular case of differentiable equations, it may work like this: the equations are elements of a vector space. In this case, then, the $dx$ in the integral $\int f(x) dx$ would be the 1-form. This becomes more intuitive when including a Jacobian transformation.
With this sketchy idea as background, I don't understand what $\omega,$ the plane through the origin, is; or why it has to go through the origin.
Perhaps I have problems understanding $\mathbb R^2$ in $T_p \mathbb R^2$ among other concepts. I see that this is the tangent plane at point $p.$ Here is a relevant passage:
Let’s look at the tangent line to the graph of $y = x^2$ at the point $(1, 1).$ We are no longer thinking of this tangent line as lying in the same plane that the graph does. Rather, it lies in $T_{(1,1)}\mathbb R^2.$ The horizontal axis for $T_{(1,1)}\mathbb R^2$ is the “dx” axis and the vertical axis is the “dy” axis. Hence, we can write the equation of the tangent line as $dy = 2dx.$
Maybe the story from a generic curved space perspective will give you a better understanding of the notation (the $\mathbb R^n$ case is somewhat special).
1-forms on a Sphere
Suppose you are in $\mathbb R^3$. The two dimensional sphere $S$ in $\mathbb R^3$ is a two dimensional manifold (manifold = generic curved space). At any point $p\in S$ the sphere has a tangent plane which we denote with $T_pS$. The subset $T_pS\subset \mathbb R^3$ is a two dimensional affine subspace of $\mathbb R^3$ and can be given the structure of a vector space in a canonical way by $q\mapsto q-p$ (i.e. identifying it with the two dimensional vector subspace of $\mathbb R^3$ that is parallel to $T_pS$). The disjoint union of the $T_pS$ is the tangent bundle of $S$, i.e. $TS := \{(p, v):\ p\in S, v\in T_pS\}$.
Now that you have a vector space for every point of $S$ you can go crazy with your linear algebra and build all sorts of other vector spaces attached to every point of $S$. One such is to consider the dual $T_pS^*$ for every $p\in S$. Once you have applied your construction at every point you take the disjoint union and get the "bundle", for example the cotangent bundle $TS^* = \{(p, a):\ p\in S, a\in T_pS^*\}$.
Every time you have a bundle like $TS^*$ you also have its projection $\pi: TS^*\to S$ that projects the elements of the bundle to the base point $p$ on $S$ that they are attached to, i.e. $\pi: (p, a)\mapsto p$. The counter image $\pi^{-1}(p) = \{p\}\times T_pS^*$ is also called the fiber of $TS^*$ at $p$.
A $1$-form on $S$ is a so called section of $\pi:TS^*\to S$, meaning a function $\alpha: S\to TS^*$ such that $\alpha(p) = (p, a_p)$ for some $a_p\in T_pS^*$, or in other words a function $S\to TS^*$ that attaches to every point $p\in S$ an element of the fiber at $p$.
1-forms on a plane
If instead of a sphere $S$ we take the plane $P = \mathbb R^2$, then for any $p\in P=\mathbb R^2$ the tangent space at $p$, $T_pP$, is the plane $P$ itself (that is why I started with the sphere, because there you can imagine a clear distinction between the tangent space and the manifold itself, unlike in this case), but with a different vector space structure (i.e. the one given by the translation $q \mapsto q - p$ where $p$ plays the role of $0$).
As before you can build $TP^* = \{(p, a):\ p\in P, a\in T_pP^*\}$. The translation $q\mapsto q - p$ gives you a canonical identification of $T_pP$ with $\mathbb R^2$, so you can use the canonical base $e_x=(1, 0), e_y = (0, 1)$ of $\mathbb R^2$ as a base for all $T_pP$ whatever $p$ is. This, in turn, gives you a base for every dual space $T_pP^*$. The vectors of this dual base are usually denoted by $dx_p, dy_p$, i.e. identifying $T_pP$ with $\mathbb R^2$, $dx_p$ and $dy_p$ are the linear functions $T_pP\to\mathbb R$ such that $dx_p(e_x) = 1, dx_p(e_y) = 0$ and $dy_p(e_x) = 0, dy_p(e_y) = 1$.
A 1 differential form on the plane $P$ is then just a function $\alpha: P\to TP$ of the form $\alpha: p\mapsto f(p) dx_p + g(p) dy_p$ for any two differentiable functions $f,g: \mathbb R^2 \to \mathbb R$ (to be pedantic it should be a function $p\mapsto (p, f(p) dx_p + g(p) dy_p)$, but that is a distinction that is almost never made in practice, as it's clear from the context that $f(p) dx_p + g(p) dy_p$ belongs to the fiber on top of $p$).
1-forms for curves in a plane
If you have a curve $\gamma: \mathbb R\to P$, you can define a differential form on it by choosing at each point $\gamma(t)$ an element of $T_{\gamma(t)}P$, i.e. a 1-form on a curve in a plane is a function of the form $t \mapsto f(t)dx_{\gamma(t)} + g(t)dy_{\gamma(t)}$.
If you fix a time $t_0$, and so a point $p_0 = \gamma(t_0)$ on the curve, the value of the differential form at $t_0$ is $\omega = f(t_0)dx_{p_0} + g(t_0)dy_{p_0}$, which is a linear function $T_{p_0}P\to \mathbb R$, and so the graph of $\omega$ is a plane through the origin in $T_{p_0}P \times \mathbb R\cong \mathbb R^2 \times \mathbb R$.