I'm thinking about the kernel of a differential one-form $\theta\in\Lambda^{1}(M)$:
$$ Ker(\theta):=\left\{X\in\mathfrak{X}(M) \;|\; \theta(X)=0\right\} $$
Now suppose $X\in Ker(\theta)$, then is $fX$, with $f$ a function on $M$, in $Ker(\theta)$?
Somewhat naively I think it is, in fact:
$$ \theta(fX)=f\theta(X)=0 $$
Is it right?
More specifically is $Ker(\theta)$ a real vector space or a real module over the functions on $M$?
My question arises because I need to understand the relationship between the foliation associated to $\theta$ and $Ker(\theta)$.
Yes, you are correct: a 1-form is $C^\infty(M)$ linear so if $\theta(X) = 0$ then $\theta(fX) = f\theta(X) = 0$ for all smooth functions $f$. This means that $\ker\theta$ is a module over smooth functions on $M$.
If $\theta$ is nowhere vanishing, then $\ker \theta$ will be a vector subbundle (i.e. distribution) of $TM$. If $\theta$ has some zeros then the rank will jump and so will not form a vector bundle but will be a sheaf of $C^\infty(M)$ modules.