Associated graded quotients are well defined subquotients

Choose the specified representatives $i_p:M_p\hookrightarrow A$. Subobject containment gives $j_p:M_{p-1}\to M_p$ with $i_p{j}_p=i_{p-1}$. If $j_{p}a=j_{p}b$, then $i_{p-1}a=i_{p-1}b$, hence $a=b$. Thus $j_p$ is monic and its cokernel exists, giving the subquotient in [F1].

For a decreasing filtration choose representatives $i^p:M^p\hookrightarrow A$. The containment $F^{p+1}A\subseteq F^{p}A$ gives $j^p:M^{p+1}\to M^p$ with $i^p{j}^p=i^{p+1}$. The same monic-cancellation argument as in step 1.1 makes $j^p$ monic, so its cokernel is the decreasing associated-graded subquotient. Equivalently, this is step 1.1 after translating by $F^{p}A=F_{-p}A$.

Replace either system of representatives by a compatible primed system and let $u$ be the compatible isomorphisms. Monicity in $A$ gives $u_p{j}_p=j_p^{\mathrm{\prime }}{u}_{p-1}$ in the increasing case and $u^p{j}^p=(j^{\mathrm{\prime }}{)}^p{u}^{p+1}$ in the decreasing case. By [F2], $u$ and $u^{-1}$ descend to maps of the respective cokernels. Their composites are identities because they agree with the identities after the epic quotient projections; the same cancellation proves uniqueness. This covers equal adjacent pieces and zero pieces in both conventions.