When is a quotient of an $R$-module $E$ a submodule of $E$? Let $R$ be a commutative ring with $1$. Suppose we are given a surjective $R$-module map $\varphi:E \to M \simeq E/N$. Are there any sufficient and/or necessary conditions for having an injective map $M \hookrightarrow E$ when $M$ is not isomorphic to a direct summand of $E$?  
Clearly this the case if there is a non-split short exact sequence
$$
0 \to M \to E \to M \to 0
$$
i.e. if $E$ is an extension of $M$ by $M$ which represent a non-zero element of $\operatorname{Ext}^1_R(M,M)$, but I can't seem to be able to get farther than this.
As an example, consider the sequence
$$
0 \to \Bbb Z/4\Bbb Z \to \Bbb Z/8\Bbb Z \oplus \Bbb Z/2\Bbb Z \to \Bbb Z/4\Bbb Z \to 0
$$
where the injection sends $1 \mapsto (2,1)$.
 A: I definitely have an idea about the special case when $E=R$, which may lead you somewhere.
A ring $R$ is called right Kasch if every simple right $R$ module is isomorphic to a right ideal of $R$. If $I$ is an essential maximal right ideal in such a ring, then $I\to R\to R/I$ can't split, and yet $R/I$ must embed into $R$.
Commutative local Artinian rings are Kasch, and so $R/M$ is isomorphic to a submodule of $R$, but since $R$ is local the submodule is not a summand.
There is another condition which I wondered if you are aware of. A module $M$ is said to have condition $C_2$ if whenever a submodule $N<M$ is isomorphic to a direct summand of $M$, then $N$ is a summand of $M$. 
That means in modules which aren't $C_2$ there can postentially be copies of a submodule, one of which is a summand and the other copy is not. This makes me think the statement of your quesiton might need tweaking. Would it be fair to say you intended "$M$ is not isomorphic to a direct summand"? (Of course, if all right $R$ modules were $C_2$, that would eliminate the ambiguity, but I'm not sure which rings have this property, at present.)
