We'll prove it under bit more general setting. Let $0\to K\to R^m\to M\to 0$ be an exact sequence where $K$ is finitely generated. Consider now $0\to L\to R^n\to M\to 0$; we wish to prove that $L$ is finitely generated.
By Schanuel's lemma, $L\oplus R^m\cong K\oplus R^n$. Then $L$ is an epimorphic image of $K\oplus R^m$, which is finitely generated.
By the way, this result is a corollary of following more general fact;
If $0\to L\to N\to M\to 0$ is an exact sequence with $M$ a finitely presented and $N$ finitely
generated, then $L$ is finitely generated.
Indeed, consider an epimorphism $R^n\to N$ and the commutative diagram
$$
\require{AMScd}
\begin{CD}
0 @>>> L' @>>> R^n @>>> M @>>> 0\\
@. @VVV @VVV @| \\
0 @>>> L @>>> N @>>> M @>>>0
\end{CD}
$$
where the arrow $R^n\to M$ is the composition of $N\to M$ after $R^n\to N$.
Then $L'$ is finitely generated by the above argument and the map $L'\to L$ is an epimorphism by diagram chasing (or four lemma).
This can also be proved by using Snake lemma instead of Schanuel's lemma (see here).
Best Answer
Yes, this holds.
Consider the following Lemmas:
Proof: a straightforward exercise. $\square$
Proof: we begin with a surjective $R$-linear map $f : R^n \to W$, and we call the standard generators of $R_n$ $e_1, e_2, \ldots, e_n$.
Now, we select, for each $1 \leq i, j \leq n$, some $w_{ij} \in R^n$ such that $f(w_{ij}) = f(e_i) f(e_j)$. We also select $q$ such that $f(q) = 1$.
For convenience, we define the bilinear operator $\star : (R^n)^2 \to R^n$ by $e_i \star e_j = w_{ij}$.
Now let $I$ be the ideal generated by $\{(e_i \star e_j) - (e_j \star e_i) \mid 1 \leq i, j \leq n\} \cup \{((e_i \star e_j) \star e_k) - ((e_i \star e_j) \star e_k) \mid 1 \leq i, j, k \leq n\} \cup \{(e_i \star q) - e_i \mid 1 \leq i \leq n\}$. It’s easy to see from the definition that $I$ is finitely generated.
Now consider the module $Q = R^n / I$, with projection map $\pi : R^n \to Q$. We see that $Q$ is finitely presented.
We define the operator $\cdot : Q^2 \to Q$ by $\pi(a) \cdot \pi(b) = \pi(a \star b)$. It’s straightforward, if a bit tedious, to confirm this operator is well-defined. Also straightforward but tedious are verifying that $\cdot$ is bilinear and satisfies the associative and commutative properties. We can also show that $\pi(q)$ is the unit of this operation, making $Q$ a commutative ring. Verify that for all $r \in R$ and $s \in Q$, we have $r \cdot q = (r \cdot 1) \cdot q$; this confirms we have an $R$-algebra structure on $Q$ compatible with its module structure.
Finally, we note that $I \subseteq \ker f$. Therefore, we have an $R$-module homomorphism $g : Q \to W$ defined by $g \circ \pi = f$; since $f$ is surjective, so is $g$. A careful analysis of $\star$ shows us that $f(a \star b) = f(a) \cdot f(b)$ for all $a, b \in R^n$. Therefore, we see that for all $a, b \in Q$, we have $g(a \cdot b) = g(a) \cdot g(b)$. So $g$ is a ring homomorphism and an $R$-linear map, hence an $R$-algebra homomorphism. $\square$
Proof: another fairly straightforward one, which I’ll leave as an exercise. Hint: if we have two surjections $f : R[X_1, \ldots, X_n] \to A$ and $g : R[Y_1, \ldots, Y_m] \to A$, we can consider a surjection $h : R[X_1, \ldots, X_n, Y_1, \ldots, Y_n]$. Show that $f$ has a finitely generated kernel iff $h$ does, iff $g$ does. $\square$
Proof: straightforward. Combine the generators of $I$ as an ideal with the generators of $A$ as a module to get generators for $I$ as a module. $\square$
Now that the tedium is out of the way, let’s continue with our proof. We know that $W$ is an $R$-algebra which is finitely generated as a module, so we invoke Lemma 2. We have some $R$-algebra $Q$ which is finitely presented as a module, together with some surjective $R$-algebra map $f : Q \to W$.
Now since $Q$ is finitely generated as a module, it is certainly finitely generated as an algebra. So take some $R$-algebra surjection $g : R[X_1, \ldots, X_n] \to Q$. Then $f \circ g$ is an $R$-algebra surjection. By Lemma 3, $\ker (f \circ g)$ is a finitely generated ideal. That is, $g^{-1}(\ker f)$ is a finitely generated ideal.
Since $g$ is surjective, we see that $\ker f$ is a finitely generated ideal of $Q$. By Lemma 4, $\ker f$ is also finitely generated as a submodule of $Q$. By Lemma 1, $W$ is a finitely presented $R$-module. $\square$