Is the kernel of a differential form a subset of $M$ or of $TM$? What do we mean by the "kernel of a differential 1-form"? In particular, if $\alpha$ is a 1-form on $M$, then I understand that it takes points $p\in M$ to functionals $\alpha_p:T_pM\to\mathbb R$. So is the kernel of $\alpha$ the set of points $p\in M$ such that $\alpha_p$ is the zero map, or is $\ker\alpha$ actually the set of vectors $v$ in $TM$ such that $\alpha_p(v)=0$ for the appropriate $p$?
I ask this mostly because I would've expected $\ker\alpha$ to be the first one, but I also saw somewhere that the kernel of the 1-form is a 2-plane distribution (which doesn't make any sense if $\ker\alpha\subseteq M$, of course).
(Also, sorry if something like this has been posted before; I saw some similar-looking posts, but I'm still a bit shaky with differential forms, so I had a lot of trouble understanding them.)
 A: In this case, we're talking about some kind of generalization of the usual notion of a "kernel." Let me explain.
As you said, a differential $1$-form is a smooth map $\alpha: M \to \mathsf{T}^*M$ from a manifold $M$ to its cotangent bundle $\mathsf{T}^*M$ (it's also required to be a section, which essentially means that it has to send each point $p \in M$ to an element of the cotangent space $\mathsf{T}^*_pM$, not an element of some other cotangent space $\mathsf{T}^*_qM$). Our function $\alpha$ sends each point $p \in M$ to a cotangent vector $\alpha_p \in \mathsf{T}^*_pM$, which is just a linear map $\alpha_p: \mathsf{T}_pM \to \mathbb{R}$.
Now, think about what we need in order define a kernel in the first place, say of some function $f: A \to B$. We want $\mathsf{ker}(f)$ to be the subset of $A$ consisting of all of those elements $a \in A$ that $f$ sends to $0 \in B$. So, we need to have an element of $B$ that we can call "zero;" but this only makes sense if $B$ is some kind of "algebraic" object, e.g. a group, vector space, module, etc. The cotangent bundle $\mathsf{T}^*M$ isn't such an object, and there isn't a single element of $\mathsf{T}^*M$ that we could sensibly call "zero," so it doesn't really make sense to talk about the "kernel" of the map $\alpha: M \to \mathsf{T}^*M$.
However, what we do have is a bunch of linear maps $\{ \alpha_p: \mathsf{T}_pM \to \mathbb{R} \}_{p \in M}$; here, for each $p \in M$, both $\mathsf{T}_pM$ and $\mathbb{R}$ are vector spaces, so it does make sense to talk about the kernel of $\alpha_p$. So what we get is a bunch of kernels $\{ \mathsf{ker}(\alpha_p) \}_{p \in M}$. In the same way that the tangent bundle $\mathsf{T}M$ is a vector bundle on $M$ (it gives us the vector space $\mathsf{T}_pM$ at each point $p \in M$), we can now define the kernel $\mathsf{ker}(\alpha)$ as the vector bundle on $M$ that gives us the vector space $\ker(\alpha_p)$ at each point $p \in M$. (In fact since each of these kernels is a subspace of the corresponding tangent space, $\mathsf{ker}(\alpha_p) \subset \mathsf{T}_pM$, what we've defined is actually a vector subbundle of $\mathsf{T}M$, which is also called a distribution.)
On a final note, remember from the rank-nullity theorem from linear algebra that at each point $p \in M$, we have
$$
\mathsf{dim}(\mathsf{T}_pM) = \mathsf{dim}(\ker(\alpha_p)) + \mathsf{dim}(\mathsf{im}(\alpha_p)).
$$
So as long as $\alpha_p$ isn't identically equal to zero, then its image will be all of $\mathbb{R}$, and hence $\mathsf{dim}(\ker(\alpha_p)) = \mathsf{dim}(\mathsf{T}_pM)-1$. This is to say that $\ker(\alpha_p)$ is a hyperplane in $\mathsf{T}_pM$. Thus a nonvanishing $1$-form defines a "hyperplane field."
(Since you mentioned contact geometry, indeed this is the same notion of hyperplane field you encounter there. A contact structure is a hyperplane field satisfying some "nonintegrability" condition, and a contact form is just a choice of $1$-form $\alpha$ – provided one exists – whose kernel, in the sense we just defined, is equal to that field.)
