It's generally true over any ring that
if a module $M$ is finitely presented, $L$ is finitely generated and $f\colon L\to M$ is a surjective homomorphism, then $\ker f$ is finitely generated.
The trick is to consider a surjective homomorphism $g\colon R^n\to L$ and consider the diagram with exact rows
$$\require{AMScd}
\begin{CD}
0 @>>> \ker f\circ g @>>> R^n @>f\circ g>> M @>>> 0 \\
@. @VVV @VgVV @| @. \\
0 @>>> \ker f @>>> L @>f>> M @>>> 0
\end{CD}
$$
It is easily seen that $\ker f\circ g\to \ker f$ is surjective as well. Thus we are reduced to prove the special case when $L=R^n$ is finitely generated and free.
By assumption, there exists a surjective homomorphism $h\colon R^m\to M$ such that $\ker h$ is finitely generated. Take the pull-back $N$ of $f$ and $h$ along $M$:
$$\begin{CD}
{} @. {} @. 0 @. 0 \\
@. @. @VVV @VVV \\
{} @. {} @. \ker f @= \ker f \\
@. @. @VVV @VVV \\
0 @>>> \ker h @>>> N @>>> R^n @>>> 0 \\
@. @| @VVV @VfVV \\
0 @>>> \ker h @>>> R^m @>h>> M @>>> 0 \\
@. @. @VVV @VVV \\
{} @. {} @. 0 @. 0 \\
\end{CD}$$
Now note that $N\cong R^n\oplus\ker h\cong R^m\oplus\ker f$, so $\ker f$ is finitely generated.
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
No, you can pretty much never expect this. For instance, suppose $R=k[x]$ and let $M=N=S\cong k[x,x']$ (with one variable being the left action of $R$ and the other variable being the right action of $R$). Then $M\otimes_R N$ will be $k[x,y,z]$, with $R$ acting on the left by $x$ and on the right by $z$ (with $y$ being in the "middle" of the tensor product, the right action on $M$ and the left action on $N$ that have been identified). This is not finitely generated as an $S$-module (it is an entire polynomial ring $S[y]$).