What is a conormal vector to a domain intuitively? I read that a conormal vector of a domain is a vector that is tangential to the domain and normal to its boundary.
If we consider an open disk in $\mathbb{R}^2$ what is a conormal vector at a point on the boundary? I can't picture it at all.
 A: The terminology "conormal" refers to covectors (covector fields), also known as 1-forms. When a field is conormal, the corresponding vector field should be normal. This correspondence is given by the duality between vector fields and 1-forms.
In the Euclidean spaces there is the standard Riemannian metric ("dot product") that allows to identify the tangent and the cotangent spaces (that is usually done in multivariable calculus).
The whole lot of intuitions can be found in this picture:

Here we have a domain $U \subset \mathbb{R}^n$ represented as $U := \{x \in \mathbb{R}^n | f(x) < 0 \}$, sot that its boundary is given as the zero set of a function $f \colon \mathbb{R}^n \to \mathbb{R}$.
Tangent vectors to the (hyper)surface $S$ arise as the velocities $\dot{\gamma}$ of smooth curves $\gamma \colon ( -\varepsilon, \varepsilon) \to S$.
Using the chain rule one easily shows that
$$
\mathrm{d}f(\dot{\gamma})=0  \tag{*}
$$
(This holds for any level surface!).
The differential $\mathrm{d}f \colon T\mathbb{R}^n \to T\mathbb{R}$ can be seen as a 1-form $\mathrm{d}f \colon \mathbb{R}^n \to \mathbb{R}$ due to the natural identification $T\mathbb{R}^n \cong \mathbb{R}^n$ that is the feature of the Euclidean spaces.
Using the standard metric in $\mathbb{R}^n$ we can convert 1-forms into vectors (that are tangent to $\mathbb{R}^n$, in fact).
Thus, the gradient $\nabla f := (\mathrm{d}f)^{\sharp}$ turns out to be a normal vector to $S$ according to (*).
Putting this the other way around, $\mathrm{d}f$ is the conormal to $S$ at all $p \in S$.
One visualizes 1-forms (at a point) as distributions of (hyper)planes so that its value on a vector corresponds to the "number" of (hyper)planes passing through this vector.
