I'm given a 1-form $\alpha$ on $\mathbb{R}^n$, and asked to compute the kernel of $d\alpha$.
Since $d\alpha$ is a 2-form on $\mathbb{R}^n$, it would eat a vector field to give a 1-form, or it would eat 2 vector fields to give a function.
Would the kernel of this 2-form be the set of all vector fields $X$ such that $d\alpha(X,-)$ is identically zero, or the set of pairs of vector fields $(X,Y)$ such that $d\alpha(X,Y)=0$? Or something else?
Your first idea is the correct one. You can construct $\ker(d\alpha)$ fiberwise as follows. In the tangent space $T_p \Bbb R^n$, pick a vector $X_p \in T_p \Bbb R^n$. Then we get a linear map $$d\alpha_p(X_p, -) : T_p \Bbb R^n \longrightarrow T_p^\ast \Bbb R^n$$ by plugging in $X_p$ as the first argument of $d\alpha_p$. Then we say that $X_p$ is in the kernel of $d\alpha_p$ if $d\alpha_p(X_p, -)$ is the zero map. These fiberwise kernels collectively form the kernel $\ker(d\alpha) \subset T \Bbb R^n$ of $d\alpha$.