Suppose to the contrary. Here we consider the zero polynomial as a polynomial with degree $0$. We get $f'(x)=A(x)(x+1)^{k-1}(x+2)^{p-1} +B(x)(x+1)^{\ell-1}(x+2)^{q-1}+C(x)(x+1)^{m-1}(x+2)^{r-1}$ for linear polynomials $A,B,C$ has at least $14$ positive roots. So does $g(x)=A(x)(x+1)^{k-m}(x+2)^{p-r}+B(x)(x+1)^{\ell-m}(x+2)^{q-r}+C(x)$. We get $g'(x)=P(x)(x+1)^{k-m-1}(x+2)^{p-r-1}+Q(x)(x+1)^{\ell-m-1}(x+2)^{q-r-1}+C'(x)$ for quadratic polynomials $P,Q$ has at least $13$ positive roots. (It could happen that $P$ is quadratic when $k=m$ or $p=r$. Similar special cases determine the degree of other polynomials defined from $Q$ on.) We get $g''(x)=R(x)(x+1)^{k-m-2}(x+2)^{p-r-2}+S(x)(x+1)^{\ell-m-2}(x+2)^{q-r-2}$ for cubic polynomials $R,S$ has at least $12$ positive roots. So does $h(x)=R(x)(x+1)^{k-\ell}(x+2)^{p-q}+S(x)$. We get $h'(x)=T(x)(x+1)^{k-\ell-1}(x+2)^{p-q-1}+S'(x)$ for a quartic polynomial $T$ has at least $11$ positive roots. Taking derivatives three more times gives $h''''(x)=U(x)(x+1)^{k-\ell-4}(x+2)^{p-q-4}+0$ for a polynomial $U$ of degree at most $7$ has at least $8$ positive roots, which is impossible.

This problem is false for stupid reasons in some edge cases like $a=b=c=d=0$ or $k=l=m=p=q=r=0$. Can we impose additional conditions for which @ThE-dArK-lOrD's argument becomes fully correct?