Homomorphic image of a module

I have this pretty trivial question: if $M\subseteq N$ are $A$-modules, is it true in general that $M$ is the direct sum of $N$ and $L:= N/M$? Or, if $0\to M\xrightarrow{\alpha} N\xrightarrow{\beta} L\to 0$ is an exact sequence of $A$-modules, does $N=M\oplus L$?

I don't think so, because I never saw such a theorem; however in this case I don't understand why in Atiyah-MacDonald (proof of proposition 6.3) it is used that, given two $A$-modules $H,H'\subseteq N$, if $\alpha^{-1}(H)=\alpha^{-1}(H')$ and $\beta(H)=\beta(H')$, then $H=H'$. Thanks for any clarify

No, not every short exact sequence is split. For example, consider the exact sequence of abelian groups: $$0\to \mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\to 0$$

Now suppose we have a short exact sequence $$0\rightarrow M\xrightarrow{\alpha} N\xrightarrow{\beta} L\rightarrow 0$$ and submodules $H,H'\subseteq N$ with $\alpha^{-1}(H) = \alpha^{-1}(H')$ and $\beta(H) = \beta(H')$. In your question, you suggest that this implies $H = H'$. In fact, this is also false.

Consider the exact sequence of abelian groups: $$0\to \mathbb{Z}\xrightarrow{\delta} \mathbb{Z}\oplus\mathbb{Z}\xrightarrow{\mu} \mathbb{Z}\to 0$$ where $\delta(n) = (n,n)$ and $\mu(a,b) = a-b$. Let $H = \mathbb{Z}\oplus \{0\}$, and let $H' = \{0\}\oplus \mathbb{Z}$. Then $\mu(H) = \mu(H') = \mathbb{Z}$, and $\delta^{-1}(H) = \delta^{-1}(H') = \{0\}$, but $H\neq H'$.

What is true is that if $H'\subseteq H$ and $\beta(H) = \beta(H')$ and $\alpha^{-1}(H) = \alpha^{-1}(H')$, then $H = H'$. This is what Atiyah and MacDonald are using in the proof of Proposition 6.3.

To see this, let $x\in H$. Since $\beta(H) = \beta(H')$, there is some $x'\in H'$ such that $\beta(x) = \beta(x')$. Thus $0 = \beta(x) - \beta(x') = \beta(x-x')$, so $x-x'\in \ker(\beta) = \mathrm{im}(\alpha)$. Let $y\in M$ with $\alpha(y) = x-x'$. Note also that $x-x'\in H$, so $y\in \alpha^{-1}(H) = \alpha^{-1}(H')$. Thus $x-x' = \alpha(y)\in H'$. So $x = (x-x') + x' \in H'$, as was to be shown.

For a fancier proof, or to put the result in context, note that we have a commutative diagram: $\require{AMScd}$ \begin{CD} 0 @>>> \alpha^{-1}(H') @>>> H'@>>> \beta(H') @>>> 0\\ @V{=}VV @V{=}VV @V{i}VV @V{=}VV @V{=}VV\\ 0 @>>> \alpha^{-1}(H) @>>> H @>>> \beta(H) @>>> 0 \end{CD} The rows are exact and the vertical arrows, except possibly for the inclusion $i\colon H\to H'$, are isomorphisms. It follows from the Five Lemma (or its special case the Short Five Lemma) that the inclusion is an isomorphism, so $H = H'$. Of course, the proof of the Five Lemma includes reasoning like the direct proof I gave above.

Atiyah and MacDonald do not explicitly state the Five Lemma, but they give a more general statement (leaving most of the proof to the reader) in Proposition 2.10. In our case, the exact sequence from Proposition 2.10 is: $$0\to \ker(=)\to \ker(i)\to \ker(=)\to \mathrm{coker}(=)\to \mathrm{coker}(i)\to \mathrm{coker}(=)\to 0$$ Now $\ker(=) = \mathrm{coker}(=) = 0$, so $\ker(i)$ and $\mathrm{coker}(i)$ are both sandwiched between $0$s in an exact sequence. It follows that $i$ is an isomorphism.

As it has already been mentioned in the comments, not every short exact sequence of modules splits.

The second statement you have is also incorrect, even when the sequence splits. Consider a field $k$ and the sequence $$0\to k\to k\oplus k\to k\to 0,$$

where the first map embeds $k$ as the first factor and the second map projects to the second factor. Then your statement simplifies to saying that there is a unique non-vertical and non-horizontal line on a plane, which is evidently not true (unless $k$ is $\mathbb{F}_2$.)

But this is not what is used in the proof of proposition 6.3! The correct statement is that if $A\subset B\subset N$, $A\neq B$, then either $\alpha^{-1}(A)\neq \alpha^{-1}(B),$ or $\beta(A)\neq \beta(B).$ And this statement is indeed true: let $x\in B\setminus A,$ if $\beta (x)\in \beta(A),$ then there exists a $y\in A$ such that $\beta(x-y)=0.$ By exactness then $x-y\in M$, but evidently $x-y\in B\cap M$, $x-y\notin A\cap M,$ since $x\notin A.$