If $F_{\bullet}R$ is a filtration of a ring $R$, the associated graded ring of $R$ is defined as $$ \mathrm{gr}_{\bullet}(R):=\bigoplus_{i \in \mathbb{N}_{0}} \mathrm{gr}_{i}(R), $$ where $\mathrm{gr}_{i}(R):=F_{i}R/F_{i-1}R$, and we set $\mathrm{gr}_{0}(R):=F_{0}R$. Multiplication of the elements in the ring $\mathrm{gr}_{\bullet}(R)$ is defined by $$ (a+F_{i}R)\cdot (b+F_{j}R):=ab+F_{i+j-1}R. $$ If $M$ is an $R$-module, we define the associated graded $\mathrm{gr}_{\bullet}(R)$-module $\mathrm{gr}_{\bullet}(M)$ as $$ \mathrm{gr}_{\bullet}(M):=\bigoplus_{j \in \mathbb{Z}} \mathrm{gr}_{j}(M), $$ where $\mathrm{gr}_{j}(M)=F_{j}M/F_{j-1}$. Here we define multiplication as $$ (a+F_{i}R)\cdot (x+F_{j}M):=ax F_{i+j}M. $$
Question: Suppose that we're given $F_{m}R \cdot F_{n}R = F_{m+n}R$ for all $m, n \in \mathbb{N}_{0}$, how can you show that $\mathrm{gr}_{n}(R)\cdot \mathrm{gr}_{m}(R) = \mathrm{gr}_{m+n}(R)$? The problem really is that to show that some $a \in \mathrm{gr}_{m+n}(R)$ is also in $\mathrm{gr}_{m}(R) \cdot \mathrm{gr}_{n}(R)$. I've found that the way multiplication is defined (only it seems for homogeneous elements) isn't that useful and I'm having trouble linking an arbitrary element of $\mathrm{gr}_{m+n}(R)$ with those in $\mathrm{gr}_{m}(R)\cdot \mathrm{gr}_{n}(R)$.
I've run in to a similar problem when trying to prove that if given $F_{m}R \cdot F_{n}M = F_{m+n}M$ for all $m\ge 0$ and $n\ge n_0$ with $n_0 \in \mathbb{Z}$ fixed, then $\mathrm{gr}_{m+n}(M)=\mathrm{gr}_{m}(R)\cdot \mathrm{gr}_{n}(M)$.
So, let's be careful. To save myself some typing I'm going to write $G_n$ instead of $\text{gr}_n$ for the associated graded.
Given some $\overline{a} \in G_{m+n}(R)$, it lifts to some $a \in F_{m+n}(R)$. By assumption it can be written $a = \sum a_m b_n$ where $a_m \in F_m(R), b_n \in F_n(R)$. Now consider the images $\overline{a_m} \in G_m(R), \overline{b_n} \in G_n(R)$ and the corresponding sum of products
$$s = \sum \overline{a_m} \overline{b_n} \in G_{m+n}(R).$$
We have
$$a - s = \overline{\sum a_m b_n} - \sum \overline{a_m} \overline{b_n}$$
so let's consider this difference. $\overline{(-)}$, as an operation $F_k(R) \to G_k(R)$, is at the very least linear, so this reduces to a sum
$$\sum \left( \overline{a_m b_n} - \overline{a_m} \overline{b_n} \right)$$
Now the multiplication $G_m(R) \times G_n(R) \to G_{m+n}(R)$ is defined so that $\overline{a_m b_n} = \overline{a_m} \overline{a_n}$ (this does not imply that $\overline{(-)}$ is a ring homomorphism, it matters that we're only considering elements of fixed degrees here) so this sum is a sum of terms equal to zero and hence is zero. So $a = s \in G_m(R) G_n(R)$ as desired. The more general result you want about acting on a filtered module has exactly the same proof.