If $0 \rightarrow M' \rightarrow M \xrightarrow[]{f} M'' \rightarrow 0$ is exact then $M$ is Noetherian iff $M',M''$ are. 
If $0 \rightarrow M' \rightarrow M \xrightarrow[]{f} M'' \rightarrow 0$ is exact then $M$ is Noetherian iff $M',M''$ are.

For an infinite chain $M_i$ in $M$ we have chains $M'_i = M_i \cap M'$ and $ M''_i = f(M_i)$ in $M',M''$, respectively. 
If both $M', M''$ are Noetherian then eventually both induced chains will halt; then w.l.o.g. $M'_i, M''_i$ are constant. If $x_i \in M_i \setminus M_{i-1}$ intersects $M'$ then $M'_i \supsetneq M'_{i-1}$, a contradiction...
That's as far as I've gotten. It's not explained in detail in Atiyah-MacDonald.

The case when $M$ is noetherian is very easy. Any ascending chain in $M'$ is also an ascending chain in $M$, so it stabilizes; for an ascending chain in $M''$, consider the ascending chain in $M$ consisting of the inverse images.
The converse implication is a bit harder. Let $L_0\subseteq L_1\subseteq\dots\subseteq L_n\subseteq \dotsb$ be an ascending chain in $M$.
Then, by assumption, the chain $L_0\cap M'\subseteq L_1\cap M'\subseteq\dots\subseteq L_n\cap M'\subseteq \dotsb$ stabilizes. Say that
$$
L_n\cap M'=L_k\cap M'\text{, for all $n\ge k$}
$$
On the other hand, the chain $f(L_0)\subseteq f(L_1)\subseteq\dots\subseteq f(L_n)\subseteq \dotsb$ stabilizes as well, so we can say that
$$
f(L_n)=f(L_k)\text{, for all $n\ge k$}
$$
(as it's not restrictive to use the same $k$ for both chains, just take the largest index from which both chains are stable).
Our task is now to show that $L_n=L_k$, for all $n\ge k$. Take $x\in L_n$; then $f(x)\in f(L_n)=f(L_k)$, so we can write $f(x)=f(y)$, for some $y\in L_k$. Now $x-y\in M'$, so $x-y\in L_n\cap M'$ and therefore $x-y\in L_k\cap M'$. Thus
$$
x=(x-y)+y\in L_k
$$
