The following is taken from: $\textit{Partial Differential Control Theory Vo 1: Mathematical tools}$ by J F. Pommaret
$\color{Green}{Background:}$
$\textbf{Definition 1.49.}$ If $M$ is a module over a ring $A,$ a $\textit{system of generators}$ of $M$ over $A$ is a family $\{x_i\}_{i\in I}$ of elements of $M$ such that any element $x\in M$ can be written $x=\sum_{i\in I}a_ix_i$ with $a_i\in A$ and $a_i=0$ for almost all $i,$ that is for all $i$ with a finite number of exceptions. In that case, we shall use to write $M=(\{x_i\}_{i\in I}).$ In particular, $M$ is $\textit{finitely generated}$ if it has a finite system of generators and $\textit{monogenic}$ if it can be generated by a single element.
$\textbf{Definition 1.50.}$ $M$ is call a $\textit{free module}$ if it has a $\textit{basis},$ that is a system of generators linearly independent over $A.$ When $M$ is free, the number of generators in a basis is called the $\textit{rank}$ of $M$ and does not depend on the basis if it is finite. Indeed, if $\{x_i\}$ and $\{y_j\}$ are two bases and $y_j=\sum {a^i}_jx_i,$ then $\text{det}({a^i}_j)$ must be invertible in $A.$ If $M$ is free of rank $n,$ then $M\cong A^n.$
$\textbf{Proposition 1.55.}$ If $A$ is a noetherian ring and $M$ is a finitely generated $A-$module, then $M$ is a noetherian module.
$\textbf{Remark 1.56.}$ In the situation of the preceding proposition, we notice that the kernel of the projection $A^n\to M$ is also a noetherian module according to the preceding lemma and thus finitely generated too. Accordingly, we may find an exact sequence of the type:
$$A^m\to A^n\to M\to 0.$$
In this case we say that $M$ is $\textit{finitely presented}$ or $\textit{presentable}$ if we do not specify the presentation. In general, a finitely generated module $M$ is said to be $\textit{coherent}$ if the kernel of an arbitrary homomorphism $A^n\to M$ is also finitely generated.
$\textbf{Proposition 1.59.}$ Let
$$0\to M'\xrightarrow{f}M\xrightarrow{g}M''\to 0$$
be a short exact sequence of $A-$modules.
$(i)$ If $M'$ and $M''$ are finitely generated, then $M$ is also finitely generated.
$(ii)$If $M$ is finitely presented and $M'$ is finitely generated, then $M''$ is also finitely presented.
$\textit{Proof.}$ We have the following commutative and exact diagram:
The left vertical column is a finite presentation of $M$ where $F$ is a free module and $K$ is finitely generated. The kernel $L$ of the composition $F\to M\to M''$ is containing $K$ as a submodule and by chasing we obtain an isomorphism $M'\equiv L/K.$ Accordingly, $L/K$ is finitely generated and $L$ is finitely generated too because of $i$). It follows that $M''$ is finitely presented.
$\color{Red}{Questions:}$
In the proof above for part $(ii)$ of Proposition 1.59, the modules $K$ and $L$ stands respectively for $\text{Ker } K,$ and $\text{Ker } L?$ Meaning the maps: $K\to F$ and $L\to F$ denote respectively $\text{Ker } K\to F$ and $\text{Ker } L\to F.$ The reason I am asking is I am not also sure the relation of modules $K, L, F$ to maps $f, g$.
Thank you in advance