$k[x]/(x^n)$ module with finite free resolution is free

Here's a more elementary explanation (from Exercise 19.7 in Eisenbud):

Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free.

Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} \ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus cannot be free (being non-faithful).

To apply this, suppose $M$ had a minimal free resolution which was finite of length $\ge 1$:

$$0 \to R^{n_m} \xrightarrow{\phi_m} R^{n_{m-1}} \to \ldots \to R^{n_0} \to M \to 0$$

By minimality, $\phi_m(R^{n_m}) \subseteq PR^{n_{m-1}}$, but $\phi_m$ is injective, so $\phi_m(R^{n_m}) \cong R^{n_m}$ is free, contradicting the proposition. Thus the finite free resolution must have length $0$, i.e. $0 \to R^{n_0} \to M \to 0$, so $M \cong R^{n_0}$ is free.

$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $\operatorname{pd}_RM<\infty$. Now we can apply the Auslander-Buchsbaum formula and get $\operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.

You needn't assume the resolution is of finitely generated free modules. You only need to assume it is free and finite, in the sense it is eventually $0$.

$R$ is an injective $R$-module. Indeed, any ideal of $R$ is of the form $(\bar x^k)$ for $k=0,\ldots,n$ since every element of $R$ can be written as $\mu \bar x^k$ for $\mu$ a unit. Suppose we're given $\eta:(\bar x^k)\to R$, and suppose $\eta(\bar x^k)=\mu \bar x^j$. Multiplication by $\bar x^{n-k}$ yields that $n-k+j\geqslant n$ so $j\geqslant k$, say $j=k+l$. Define $\tilde\eta:R\to R$ by $\bar 1\to \mu \bar x^l$. Then $\tilde\eta$ extends $\eta$. By Baer's criterion, $R$ is $R$-injective. Since $R$ is local, projective and free modules coincide, and since $R$ is noetherian the theorem of Bass Papp says any direct sum of copies of $R$ is injective, that is any projective $R$-module is injective. But then I claim any module that admits a finite projective resolution must itself be projective. Indeed, say we have a module that admits a resolution $$0\to P_n\to P_{n-1}\to \cdots \to P_0\to M\to 0$$ but no shorter resolution. We can assume $n=1$, since $M'=P_{n-1}/P_{n}$ admits $0\to P_{n-1}\to P_n\to M'\to 0$ but cannot itself be projective, else we would be able to shorten $M$'s resolution to one of the form $$0\to M'\to P_{n-2}\to\cdots P_0\to M\to 0$$ contrary to our assumption. Then for $M'$ we have a SEC $$0\to P_1'\to P_0'\to M'\to 0$$ which cannot split. But it does, since $P_1'$ being projective is injective. The conclusion is that any $R$-module with a finite free resolution must in fact be free.