First we offer clarification on the statement of the theorem.
Suppose that $R$ is a (not necessarily commutative) associative ring $L, M, N\in R-\text{mod}$ (left $R$-modules, as without commutativity we must pick a side), and further suppose that $g:L\to M$ and $h:M\to N$ are $R$-module homomorphisms. For any $R$-module $D$, we have a functor $\hom_R(D,-):R-\text{mod}\to \mathbb Z-\text{mod}$, defined on morphisms by composition: if $g:L\to M$, then $g_*:\hom_R(D,L)\to \hom_R(D,M)$ is defined by $g_*(f)=g\circ f$.
It is not too difficult to show that if $0\stackrel{}{\longrightarrow} L\stackrel{g}{\longrightarrow} M\stackrel{h}{\longrightarrow} N$ is a left exact sequence, then $0\longrightarrow \hom_R(D,L)\stackrel{g_*}{\longrightarrow} \hom_R(D,M)\stackrel{h_*}{\longrightarrow} \hom_R(D,N)$ is also left exact.
The theorem is the converse of this statement. If $L, M, N, g, h$ are all given, and if for every choice of $D$ the sequence $0\longrightarrow \hom_R(D,L)\stackrel{g_*}{\longrightarrow} \hom_R(D,M)\stackrel{h_*}{\longrightarrow} \hom_R(D,N)$ is left exact, then $0\stackrel{}{\longrightarrow} L\stackrel{g}{\longrightarrow} M\stackrel{h}{\longrightarrow} N$ is left exact.
There is NOTHING to prove about things being maps of $R$-modules, because things are assumed to be maps of $R$-modules. The only things to prove are about exactness.
There isn't much to prove here, because you can simply take $D=R$, and there is a cannonical isomorphism $\hom_R(R,M)\cong M$ $f\mapsto f(1)$. There is a corresponding theorem with the contravariant functors $\hom_R(-,D)$ which is more interesting.
However, it is worth spending a bit of time generalizing the fact that when $D=R$, we are getting not just abelian groups but $R$ modules.
Suppose that $D$ is an $(R,S)$-module (left $R$-module, right $S$-module, and the two structures commute with each other, i.e., $(rd)s=r(ds)$), then the hom-sets will carry more structure and the induced morphisms will be compatible with that structure.
If $D\in (R,S)-\text{mod}$ and $M\in R-\text{mod}$, then we can define multiplication $s\cdot f$ on $f\in \hom_R(D,M)$ by $(s\cdot f)(d)=f(ds)$. Then $[(st)\cdot f](d)=f(dst)$ while $[s\cdot (t\cdot f)](d)=(t\cdot f)(ds)=f(dst)$
If $M$ and $N$ are $R$-modules and $g:M\to N$ is an $R$-module homomorphism, then the induced map is a map of $S$-modules. We need to verify $g_*(s\cdot f)=s\cdot (g_*(f))$. For every $d\in D$ we can evaluate and then unravel the definitions. $g_*(s\cdot f)(d)=g\circ (s\cdot f)(d)=g(f(ds))=s\cdot (g(f(d))=[s\cdot (g_*(f))](d)$.
It is also worth remarking that if the map is not a map of $R$-modules, then the induced map will not exist (or rather, will not go between the required spaces). Let $M, N\in R-\text{mod}$, and $g:M\to N$ be a map of abelain groups which is NOT a map of $R$-modules, so that there exists $m, r$ such that $g(rm)\neq rg(m)$. If $f\in \hom_R(D,M)$, then $g\circ f$ will not in general be an $R$-module homomorphism. Suppose that $f(d)=m$. Then $$r(g_*(f))(d)=r[(g\circ f) (d)]=r(g(f(d)))=r(g(m))\neq g(rm)=g(rf(d))=[g_*(rf)](d).$$
Best Answer
I can give some reasons why someone working with modules ought to care about these properties and abstract them into definitions.
For flat modules, the definition arises naturally when we intuit something from the notation, and then discover that this notation can sometimes be misleading.
For injective and projective modules, the definition comes from looking at how homomorphisms naturally arise from other homomorphisms in certain situations, and asking the question "Does every homomorphism arise in this natural way?"
Flat:
Let $P, M, N$ be modules over your commutative ring $R$. Suppose that $M$ is a submodule of $N$. The generators of $N \otimes_R P$ are written as $n \otimes p$ for $n \in N$ and $p \in P$ and likewise for the generators of $M \otimes_R P$.
There is a natural $R$-module homomorphism $\varphi: M \otimes_R P \rightarrow N \otimes_R P$ given on generators by
$$\varphi(m \otimes p) = m \otimes p$$
By the notation, it looks like $\varphi$ doesn't actually do anything. From the notation, and since $M$ is a subset of $N$, it looks like $M \otimes_R P$ should be a submodule of $N \otimes_R P$. But this isn't always true. You cannot guarantee this for all inclusions of submodules $M \subset N$ unless $P$ is flat. This is one possible definition you can take for flat modules: modules for which the above intuition always works.
Injective:
Easier to motivate this for abelian groups (that is, $\mathbb Z$-modules). Let $A$ be a subgroup of an abelian group $B$. Let $\mathbb C^{\ast}$ be the (multiplicative) group of complex numbers. If $\chi: A \rightarrow \mathbb C^{\ast}$ is a group homomorphism, one might wonder whether it is possible to extend $\chi$ (possibly nonuniquely) to a group homomorphism $\overline{\chi}: B \rightarrow \mathbb C^{\ast}$. It turns out that this is indeed possible, because $\mathbb C^{\ast}$ is an injective object in the category of abelian groups.
You can take this as the definition of injective: an abelian group $C$ is injective if whenever $A \subset B$ are abelian groups, every homomorphism of $A$ into $C$ can be extended (possibly nonuniquely) to a homomorphism of $B$ into $C$.
This is a pretty useful property for abelian groups to have. I ended up randomly needing the fact that $\mathbb C^{\ast}$ is an injective abelian group in my dissertation.
Projective:
Let $R$ be a commutative ring with identity, and let $I \subset J$ be ideals of $R$. There is a natural ring homomorphism
$$R/I \rightarrow R/J$$
$$r+I \mapsto r+J$$
Now let $B$ be another commutative ring with identity. If you're given a homomorphism of $B$ into $R/I$, you can compose with the natural homomorphism above and get a homomorphism of $B$ into $R/J$. It's natural to ask whether there are homomorphisms of $B$ into $R/J$ which don't arise in this fashion. This is the sort of idea that projective modules are based on.
Let $P$ be an $R$-module, and let $N$ be a submodule of an $R$-module $M$. You have a natural $R$-module homomorphism $\pi: M \rightarrow M/N$. If $\varphi: P \rightarrow M/N$ is an $R$-module homomorphism, you might ask the question of whether $\varphi$ always comes from a (possibly nonunique) $R$-module homomorphism of $P$ into $M$. The answer "Yes for all choices of $N \subset M$" is equivalent to $P$ being projective.