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,$$i.e., the limit of f_t f_t=\partial f/ \partial t also has an A_4 at least. Now using f_x=0, f_y=0 and the vanishing (in the limit) of f_{ij} for i+2j\le 4 which you already know, it is possible to write the unknowns f_{40t}(x,y,0), f_{t40}(0,0,0), f_{02t}(x,y,0), f_{t02}(0,0,0), f_{21t}(x,y,0) f_{t21}(0,0,0) in terms of f_{50}, f_{31} and f_{12}. Substitute in Q, and the resulting equation is exactly what you were looking for. 1 Almost everything in this answer has already been said by qui-vadis or in the comments, but now I'll translate it to your notation. I'll write f(x,y,t) for f, and f(x,y,0) for the limit f. First remark that u^4(u-t)^2 divides f(L_1u,u^2,t) so u^6 divides f(L_1u,u^2,0) (this is qui-vadis' local Bézout). Expanding this and passing to the limit,$$\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,$$i.e., the limit of$f_t$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$f_{40t}(x,y,0)$,$f_{02t}(x,y,0)$,$f_{21t}(x,y,0)$in terms of$f_{50}$,$f_{31}$and$f_{12}$. Substitute in$Q\$, and the resulting equation is exactly what you were looking for.