The equivalent proposition is:
Proposition. $R$ and $S$ are Morita equivalent if and only if there is a progenerator $P$ of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$.
While searching for the answer of this question, I found some texts in the internet which stated the above proposition putting $\text{End}(P) \simeq S$ instead of $\text{End}(P) \simeq S^{\text{op}}$, and it is wrong! I thought it was important to share this detail, because it may confuse people, as it confused me. Now let's see the explanation:
The fundamental motive for the appearence of $S^{\text{op}}$ instead of $S$ is the way we compose functions. We have the following (which is easily verified):
If we regard the ring $S$ as a right module ($S_{S}$), then $\text{End}(S_{S}) \simeq S$ and if we regard $S$ as a left module ($_{S}S$), then $\text{End}(_{S}S) \simeq S^{\text{op}}$ (those are ring isomorphisms).
Now, for more details, let us examine the necessity of those propositions closely:
If $F : \text{Mod}_{S} \rightarrow \text{Mod}_{R}$ is an equivalence of categories, then putting $P_{R} = F(S_{S})$, we may show that $P_{R}$ is a progenerator, and to verify that $\text{End}(P_{R}) \simeq S$, we proceed like this: $$\text{End}(P_{R}) = \text{End}(F(S_{S})) \simeq \text{End}(S_{S}) \simeq S.$$
And for the left case we have:
If $F : \text{ }_{S}\text{Mod} \rightarrow \text{ }_{R}\text{Mod}$ is an equivalence of categories, then putting $_{R}P = F(_{S}S)$, we may show that $_{R}P$ is a progenerator, and we have: $$\text{End}(_{R}P) = \text{End}(F(_{S}S)) \simeq \text{End}(_{S}S) \simeq S^{\text{op}}.$$
To prove the sufficiency of the propositions, we proceed as follows:
If $P$ is a progenerator of $\text{Mod}_{R}$ such that $\text{End}(P) \simeq S$, then we may show that $$\text{Hom}_{R}(P,-) : \text{Mod}_{R} \rightarrow \text{Mod}_{S}$$ is an equivalence of categories, and for $M$ in $\text{Mod}_{R}$, we regard $\text{Hom}_{R}(P,M)$ as a right $\text{End}(P)$-module in the usual way, by composing functions, so that it becomes a right $S$-module.
For the left case we have:
If $P$ is a progenerator of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$, then we may show that $$\text{Hom}_{R}(P,-) : \text{ }_{R}\text{Mod} \rightarrow \text{ }_{S}\text{Mod}$$ is an equivalence of categories, and for $M$ in $_{R}\text{Mod}$, we regard $\text{Hom}_{R}(P,M)$ as a left $\text{End}(P)^{\text{op}}$-module in the usual way, by composing functions and making the necessary adjustments (putting the $^{\text{op}}$) so that everything works, and then it becomes a left $S$-module, since $\text{End}(P)^{\text{op}} \simeq (S^{\text{op}})^{\text{op}} = S$.
I think your hypotheses lack some necessary precision. Is $R$ commutative? unital? Noetherian, left or right? Is $N$ a left $R$-module? I think you intended $R$ to be commutative but what I wrote below doesn't require it.
$a) \Rightarrow b)$ seems incorrect or at least incomplete. It is possible for a vector space to be a sum of three vector subspaces which aren't in direct sum, even though they are pairwise in direct sum. For instance, $K^2=K(0,1) + K(1,0) + K(1,1)$, and the lines have pairwise intersection zero, but the overall sum isn't direct.
I'd prove it this way: let $N=\sum_i{N_i}$ where $N_i$ is simple. Because $N$ is finitely generated, we can assume that the $N_i$ are finitely many, and we can thus name them $N_1,\ldots,N_m$. For each $i$, call $N_i$ new if its intersection with $\sum_{j < i}{N_j}$ is zero. As $N_i$ is simple, if $N_i$ is not new then $N_i \subset \sum_{j < i}{N_j}$. It is then easy enough to check that $N$ is the direct sum of the new $N_i$.
Uniqueness in $b) \Rightarrow c)$ seems false. For instance, in $K^2$, we have $K^2=K(0,1)\oplus K(1,0)=K(0,1)\oplus K(1,1)$. But let's see if we can show existence: write a finite sum $N=\bigoplus_{i \in I}{N_i}$ where the $N_i$ are simple. Let $M$ be a submodule of $N$.
Let $J \subset I$ be a maximal subset such that $\sum_{j \in J}{N_j}$ is in direct sum with $M$. Let's show that $N=M \oplus \sum_{j \in J}{N_j}$. Clearly, it's enough to show that $P = M \oplus \sum_{j \in J}{N_j}$ contains every $N_i$. If $i \in J$, that's clear. If not, we know that $P$ isn't in direct sum with $N_i$ (by maximality of $J$), that is, $P \cap N_i$ is nonzero. As $N_i$ is simple, it follows $N_i \subset P$.
Again, how about existence in $c)$ implying $a)$?
I'll simply prove that every submodule of $N$ satisfies existence in $c)$ as well. Indeed, let $M_1 \subset M$ be submodules, let $M' \subset N$ be such that $M \oplus M'=N$. Let then $M''$ be such that $M' \oplus (M'+M_1)=N$. Let $\pi:N \rightarrow M$ be the projection with kernel $M'$. It's easy to check that $M = \pi(M'') \oplus M_1$.
Note also that every submodule of $N$ is the image of $N$ under some projection so is finitely generated. In particular, every sequence of submodules $(M_n)_n$ such that $M_n \subset M_{n+1}$ is stationary.
Let $(M_n)$ be a sequence of submodules such that $M_n \supset M_{n+1}$. Let $M_n=N_n \oplus M_{n+1}$. Then $\left(\bigoplus_{p \leq n}{N_p}\right)_n$ is a nondecreasing sequence of submodules of $N$ so is stationary, which implies that for $n$ large enough $N_n=0$, ie $M_n=M_{n+1}$ so $M_n$ is stationary.
From this, it follows that the set of nonzero submodules of $N$ has minimal elements (the simple submodules). Let $S$ be their sum, we can write $N=S \oplus S_1$ for some submodule $S_1$. Assume $S_1$ is nonzero: since every non-increasing/nondecreasing sequence of submodules of $S_1$ is stationary, $S_1$ contains a simple submodule $P$. But $P \subset S$ by definition of $S$, so $P \subset S\cap S_1 = \{0\}$, a contradiction. So $N=S$ is sum of simple modules.
Best Answer
Let $G$ denote an inverse of $F$ with $G(N_i) = A_i$. Then $G$, being exact, sends the monomorphism $N_i\hookrightarrow F(M)$ to the monomorphism $A_i\hookrightarrow G(F(M))\cong M$. Hence you can identify $A_i$ with a submodule of $M$