Let $M$ be a f.g. module over a Noetherian ring. I'm trying to fill in the details in the proof of the fact that $\mathrm{Ass}(M)$ is finite. It should follow from Proposition 3.7 and Lemma 3.6 in Eisenbud.
First, consider a filtration $0=M_0\subset\dots\subset M_n=M$ where $M_{i+1}/M_{i}\simeq R/P_i$ for some prime ideals $P_i$ (Proposition 3.7 in Eisenbud). Let us prove by induction on $n$ that $\mathrm{Ass}(M_n)\subset \{P_0,\dots, P_{n-1}\}$.
First of all, I have problems with the base case: if $n=1$, the chain looks like $0=M_0\subset M_1=M$, and $M_1\simeq M_1/M_0\simeq R/P_0$ for some prime ideal $P_0$. So $P_0$ is an associated prime of $M$, so $\{P_0\}\subset \mathrm{Ass}(M)$. Why does the reverse inclusion hold?
Assume the result holds for all $k<n$. Consider the exact sequence $0\to M_{n-1}\to M_n\to M_n/M_{n-1}\to 0$. By lemma 3.6, $$\mathrm{Ass}(M_n)\subset \mathrm{Ass}(M_{n-1})\cup \mathrm{Ass}(M_n/M_{n-1}).$$
By hypothesis $\mathrm{Ass}(M_{n-1})\subset \{P_0,\dots,P_{n-2}\}$. But what is $\mathrm{Ass}$ of the quotient, and why?
In general, if $P\subset R$ is prime, then $\mathrm{Ass}(R/P)=\{P\}$. Indeed, the annihilator of any nonzero element of $R/P$ is $P$, since $P$ is prime. (If a nonzero element $a\in R/P$ were annihilated by some $b\not\in P$, then lifting $a$ to an element $a'\in R$, we would have $a'b\in P$ but $a',b\not\in P$, a contradiction.)
This also answers your second question: $M_n/M_{n-1}\cong R/P_{n-1}$, so $\mathrm{Ass}(M_n/M_{n-1})$ is just $\{P_{n-1}\}$.