Suppose $R$ is a ring with filtration $F_{\bullet}R$: $$ \{0\} \subseteq F_0R \subseteq F_1R \subseteq \cdots \subseteq F_{n}R \subseteq \cdots \subseteq R. $$ Let $\mathrm{gr}_{\bullet}^{F}R$ be the associated graded algebra $$ \mathrm{gr}_{\bullet}^{F}R:=\bigoplus_{i=0}^{\infty} \mathrm{gr}_{i}^{F}R \quad \text{where}\; \mathrm{gr}_{i}^{F}R:=F_{i}R/F_{i-1}R \;\text{and}\; \mathrm{gr}_{0}^{F}R:=F_{0}R. $$ Let $M$ be an $R$-module with filtration $F_{\bullet}M$: $$ \cdots \subseteq F_{i}M \subseteq F_{i+1}M \subseteq \cdots \subseteq M \quad (i \in \mathbb{Z}) $$ such that $\bigcup_{j} F_{j}M=M, \bigcap_{j}F_{j}M=\{0\}$ and $F_{i}R\cdot F_{j}M\subseteq F_{i+j}M$ for all $i\ge 0$ and $j \in \mathbb{Z}$.
We form the associated graded $\mathrm{gr}_{\bullet}^{F}R$-module: $$ \mathrm{gr}_{\bullet}^{F}M:=\bigoplus_{j \in \mathbb{Z}}\mathrm{gr}_{j}^{F}M \quad \text{where}\; \mathrm{gr}_{j}^{F}M:=F_{j}M/F_{j-1}M $$ The module structure is given by $$\mathrm{gr}_{i}^{F}R \cdot \mathrm{gr}_{j}^{F}M \subseteq \mathrm{gr}_{i+j}^{F}M \quad \text{for all}\; i,j \in \mathbb{Z}.$$
Question: If $\mathrm{gr}_{\bullet}^{F}M$ is finitely generated over $\mathrm{gr}_{\bullet}^{F}R$, (1) why/how is it that we can write $$ \mathrm{gr}_{\bullet}^{F}M=\mathrm{gr}_{\bullet}^{F}R \cdot \bigoplus_{j\le j_0} \mathrm{gr}_{j}^{F}M \quad \text{for some}\; j_0 \in \mathbb{Z}?$$ and (2) how does it follow from this that:
$$\mathrm{gr}_{l+1}^{F}M = \sum_{j\le j_0} \mathrm{gr}_{l+1-j}^{F}R \cdot \mathrm{gr}_{j}^{F}M \subseteq \mathrm{gr}_{1}^{F}R \cdot \mathrm{gr}_{l}^{F}M $$ for all $l \ge j_0$?
Finally, (3) obviously the above inclusion implies that $\mathrm{gr}_{l+1}^{F}M = \mathrm{gr}_{1}^{F}M \cdot \mathrm{gr}_{l}^{F}M$, but then why does it follow from this that $$F_{l+1}R = F_1R \cdot F_l M + F_l M \subseteq F_1R \cdot F_l M?$$
I'm extremely sorry is this is too many questions. I wouldn't ask so many, but they're all very related and I think it would be best to ask them together.
EDIT: Problem (1) when I think a little hard about seems like obviously true and I don't believe requires any justification.
For problem (2), the equality in the displayed formula follows directly from above, but I don't understand how you get the inclusion.
For problem (3), from a comment here of @Mindlack, how the equality is obtained is answered. I don't know about the inclusion still.
$\newcommand{\gr}{\operatorname{gr}}$ For question (2) you have to use $\gr^F_iR\cdot \gr^F_jR = \gr^F_{i+j}R$ for all $i,j$, which follows from the (still missing) assumption that $F_iR\cdot F_jR = F_{i+j}R$ for all $i,j$. Indeed, given $l\ge j_0$ and $j\le j_0$, we then have $$ \gr^F_{l+1-j}R\cdot \gr^F_jM = \gr^F_1R\cdot \gr^F_{l-j}\cdot \gr^F_jM \subseteq \gr^F_1R\cdot \gr^F_lM. $$ Hence, taking the sum over all $l\ge j_0$ yields the claim.
Ad (3): We have $\gr^F_{l+1}M = \gr^F_1R\cdot \gr^F_lM$. By writing out how $\gr^F_\bullet R$ acts on $\gr^F_\bullet M$, you see that this is equivalent to $$ F_{l+1}M = F_1R\cdot F_lM + F_lM. $$ To be more explicit: Recall that $\gr^F_lM = F_lM/F_{l-1}M$ and similarly $\gr_1R = F_1R/F_0R$. Then $\gr^F_{l+1}M = \gr^F_1R\cdot \gr^F_lM$ means that for every $m+F_lM \in \gr^F_{l+1}M$ there exist $r_1,\dotsc,r_n\in F_1R$ and $m_1,\dotsc,m_n\in F_lM$ such that $$ m+F_lM = \sum_{i=1}^n r_im_i + F_lM. $$ But then $m - \sum_{i=1}^nr_im_i = m'\in F_lM$, i.e., $m = \sum_{i=1}^n r_im_i + m'$. This shows $F_{l+1}M \subseteq F_1R\cdot F_lM + F_lM$. The other inclusion follows from the definition of a filtration of a module.
But notice that $1 \in F_0R \subseteq F_1R$ (this is true for the order filtration and the Bernstein filtration in your source, but regardless of that it should be part of the definition of “filtration of a ring”). Hence, $F_lM \subseteq F_1R\cdot F_lM$, so that finally $$ F_{l+1}M = F_1R\cdot F_lM. $$