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

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.

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.

You can use elementary tools to show this. In here elementary methods are used assuming the modules involved are finitely generated.

Set $R=k[x]/(x^n)$. To solve your problem let us prove the following:

Let $M$ be free a $R$-module. If $B\subseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$.

As $M$ is free, there is a set $I$ such that $$ M\simeq\bigoplus_IR,$$

so we consider $M$ as this direct sum. Define the function $$\psi:M\rightarrow\bigoplus_I k,$$ by letting for each $i\in I$ and each $a\in M$ $\text{ }\psi(a)_i$ to be the constant term of $a_i$. As for each $a\in M$ we have $x^{n-1}a=x^{n-1}\psi(a)(1)$, from the linear independence of $B$ over $R$ we get that the set $\{\psi(x):x\in B\}$ is a linear independent subset of the $k$-vector space $\bigoplus_I k$.

Thus there is a basis $C$ of the $k$-vector space $\bigoplus_I k$ such that $\{\psi(x):x\in B\}\subseteq C$. We claim the set $$B'=(C\setminus \{\psi(x):x\in B\})\cup B$$ is a basis of $M$ over $R$; notice $\psi$ is $1-1$ on $B'$ and $\psi(B')=C$.

Let us see by reverse induction on $0\leq j\leq n-1$ that for each $a\in M$ with $deg(a_i)\geq j$ or $a_i=0$ for each $i\in I$ we have $a\in$Span$_R(B')$

From $(1)$ and because of the construction of $C$ the case $j=n-1$ follows.

If for some $0<j\leq n-1$ the property holds, pick any $a\in M$ with $\deg(a_i)\geq j-1$ or $a_i=0$ for all $i\in I$, then by the inductive hypothesis we may assume that for each $i\in I$, $a_i=b_ix^{j-1}$ for some $b_i\in k$. Now consider the set $\{x^{j-1}b:b\in B'\}$, by the inductive hypothesis we may assume that each $x^{j-1}b$ equals $x^{j-1}\psi(b)$, hence as $\psi(B')=C$ and $\psi$ is $1-1$ on $B'$ we get that $a\in$Span$_R(B')$.

Finally, let's check $B'$ is linearly independent over $R$. Suppose $c_1,\ldots, c_m\in B'$ are distinct and $p_1,\ldots p_m\in R$ are such that $$p_1c_1+\cdots+p_mc_m=0.$$

Each $p_i$ can be written as $p_i=\sum_{j=0}^{n-1}b_{i,j}x^j$. Let us see by induction on $0\leq j\leq n-1$ that $b_{i,j}=0$ for all $1\leq i\leq m.$

We have
$$0=x^{n-1}p_1c_1+\cdots+x^{n-1}p_mc_m=x^{n-1}a_{1,0}\psi(c_1)+\cdots+x^{n-1}a_{1,m}\psi(c_m),$$ thus as $\psi$ is $1-1$ on $B'$ and $\psi(B')=C$ we obtain $a_{1,i}=0$ for all $1\leq i\leq m.$

If for some $0\leq j<n-1$ we have that $a_{j',i}=0$ for all $0\leq j'\leq j$ for all $i$, then $$0=x^{n-2-j}p_1c_1+\cdots+x^{n-2-j}p_mc_m=x^{n-2-j}a_{j+1,0}\psi(c_1)+\cdots+x^{n-2-j}a_{j+1,m}\psi(c_m);$$

since $j+1+(n-2-j)=n-1$, and as we argued in the previous case we get $a_{j+1,i}=0$. Therefore $p_1=\cdots=p_m=0$.

Therefore $B'$ is a $R$-basis of $M$ with $B'\supseteq B$.

Now suppose $M$ is an $R$-module that has a free finite resolution

$$0\rightarrow M_m\rightarrow M_{m-1}\cdots\rightarrow M_1\rightarrow M\rightarrow 0,$$

where the map $M_j\rightarrow M_{j-1}$ is denoted by $f_j$, and $M_0=M$.

Let us see by reverse induction on $1\leq j\leq m$ that $Im(f_j)$ is a free submodule of $M_{j-1}$. The case $j=m$ is trivial as $f_m$ is injective and $M_m$ is free.

So suppose $Im(f_j)$ is free for some $1<j\leq m$. Then by what we prove above we know $$Im(f_{j-1})\simeq M_{j-1}/\ker(f_{j-1}),$$ is free since $M_{j-1}$ is free and $\ker(f_{j-1})=Im(f_j)$ is free.

Therefore $M$ is free.