Your $K^\mathrm{fin}$ is certainly closed under linear combinations with coefficients in $k$. In particular if $D_1 f_1 = 0$ and $D_2 f_2 = 0$, then the pair $(f,f_2) = (f_1+f_2,f_2)$ ($f$ could also be a different linear combination) will satisfy the system of equations $D_1 f = D_1 f_2$, $D_2 f_2 = 0$. Now, it remains to differentiate the first equation as many times as needed to eliminate $f_2$ and its derivatives. After elimination you'll have an equation $Df = 0$ in terms of $f$ and its derivatives alone. To make sure that you need to differentiate only finitely many times, every time a derivative of $f_2$ of order equal to or greater than the order of $D_2$, it can be eliminated using the equation $D_2 f_2 = 0$. So, the number of times you need to differentiate $D_1 f = D_1 f_2$ to eliminate $f_2$ is bounded by the order of $D_2$. This is a classic trick, though I don't know whether it has any name or standard reference associated to it.
Of course, you can play the same game with $(f,f_2) = (f_1 f_2,f_2)$ and the system of equations $D_1(f/f_2) = 0$, $D_2 f_2 = 0$. However, after eliminating $f_2$ and its derivatives in the naive way, you end up with a non-linear equation for $f$. It is not clear to me at the moment whether differentiating a larger number of times than one might naively expect could result in a linear equation for $f$. So I don't know the answer to the question of the closure of $K^\mathrm{fin}$ under multiplication. I've also not found a counter example.