Almost everything in this answer has already been said by qui-vadis or in the comments, but now I'll translate it to your notation. I'll write $f(x,y,t)$ for $f$, and $f(x,y,0)$ for the limit $f$.
First remark that $u^4(u-t)^2$ divides $f(L_1u,u^2,t)$ so $u^6$ divides $f(L_1u,u^2,0)$ (this is qui-vadis' local Bézout). Expanding this and passing to the limit, $$\frac{L_1^5}{5!}f_{50}+\frac{L_1^3}{3!}f_{31}+\frac{L_1}{2}f_{12}=0.$$
Next oberve that the limit vanishings of $f_{40}$, $f_{21}$ and $f_{02}$, together with $f_{40}(L_1t,t^2,t)f_{02}(L_1t,t^2,t)−3f^2_{21}(L_1t,t^2,t)=0$, give $$Q:=f_{40t}(x,y,0)f_{02t}(x,y,0)−3f^2_{21t}(x,y,0)=0,$$ $Q:=f_{t40}(0,0,0)f_{t02}(0,0,0)−3f^2_{t21}(0,0,0)=0,$$ i.e., the limit of $f_t$ f_t=\partial f/ \partial t$ also has an $A_4$ at least. Now using $f_x=0, f_y=0$ and the vanishing (in the limit) of $f_{ij}$ for $i+2j\le 4$ which you already know, it is possible to write the unknowns $f_{40t}(x,y,0)$, f_{t40}(0,0,0)$, $f_{02t}(x,y,0)$, f_{t02}(0,0,0)$, $f_{21t}(x,y,0)$ f_{t21}(0,0,0)$ in terms of $f_{50}$, $f_{31}$ and $f_{12}$. Substitute in $Q$, and the resulting equation is exactly what you were looking for.
