If $\mathcal{R}\subset J^rX$ is closed, then there's a smooth function $f:J^rX\to\mathbb R$ with $\mathcal{R}=f^{-1}(0)$. So you can construct a differential operator $H:J^kX\to M\times \mathbb{R}$ by $H(\theta):=(\pi_X(\pi^r_0(\theta)),f(\theta))$ and the equation $\mathcal{R}$ will be given by $H(j^r\phi)=0$.
So there is no big difference between the two definitions. If you are only interested in the "space" of solutions of the differential equation, then i'd say that the set $\mathcal{R}$ is enough, or put differently, you could choose the differential operator which suits you best to represent the equation.