I will address a few points of $(2)$ and provide useful reformulations.
First reformulation. A $\mathbb{Z}_n[x_1^{\pm}, \ldots, x_D^{\pm} ]$-module is the same as a $\mathbb{Z}_n$-module equipped with the action of $D$ automorphisms $x_1, \ldots, x_D$ commuting with each other.
For a matter of convenience, let us restrict to the case $n=p$ prime, so that the underlying module is an $\mathbb{F}_p$ vector space.
Second reformulation. A module $M$ is quasi-free if and only if there exists a subspace $M_{0 \ldots 0} \subset M$ such that $$ M \simeq \bigoplus_{n_1 \ldots n_d \in \mathbb{Z} } M_{n_1 \ldots n_d} $$ where $M_{n_1 \ldots n_d} := x_1^{n_1} \ldots x_D^{n_D} M_{0 \ldots 0}$.
This amounts to spelling out the action of the $x_i$'s in $M_0 \otimes_{\mathbb{Z}_n} \mathbb{Z}_n [x_1^{\pm}, \ldots x_D^{\pm} ]$
We further restrict to $D=1$, which is enough to produce relevant counterexamples. In this case, a quasi-free module is a sequence of vector spaces $\{V_n\}_{n \in \mathbb{Z}}$ with isomorphisms $ f_n: V_n \to V_{n+1}$ (a sort of infinitely-long quiver representation). We have a simple numerical criterion for quasi-freeness:
Observation Suppose $M$ is quasi-free and finitely generated over $R$. Then there exists an integer $m > 0$ such that $$\lambda_k(M) := \dim_{\mathbb{F}_p}(M/x^k)M = km$$ for all $k \ge 1$. Furthermore, $x$ acts on $M/x^kM$ as a block-diagonal matrix with blocks $\left ( \begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix} \right )$. Note that the decomposition above of $M = \bigoplus_{n \in \mathbb{Z}} M_n$ becomes $M_0 \oplus \ldots \oplus M_{k-1}$ when quotienting for $x^k$. If the latter is finite-dimensional, a basis $v_1, \ldots, v_k$ of $M_0$ as well as their images $x^i v_1, \ldots, x^iv_k, i\le k-1$ provides the desired basis for the block-diagonal representation and the dimensionality criterion.
We are ready to provide the first
Counterexample. There exists an exact sequence $0 \to A \to B \to C \to 0$ such that $A,C$ are quasi-free but $B$ is not. Proof. Consider $A_n := \mathbb{F}_pa_n, C_n := \mathbb{F}_p c_n$. Define $B_n = <a_n, c_n>$ with $f_n$ given by $$ f_n(a_n) = a_{n+1}, f_n(c_n) = c_{n+1} + a_n $$ With the intuitive maps $A \to B \to C$. It is exact since it is so as vector spaces. We want to apply the above criterion. Of course, $\lambda_1(B) = \dim_{\mathbb{F_p}}(B/xB) = 2$. On the other hand, we claim that $\lambda_2(B) \le 3$. It is easy to see that we have an isomorphism $$ B/x^2 B \simeq (B_0 \oplus B_1) / x^2(B_0 \oplus B_1) $$ The claim follows since one of the four generators $x^2(c_0) = x(c_1+a_0) =a_1$ becomes zero in the quotient.
I would like to come back and provide other examples along this line (or, hopefully, prove some stability results for quasi-freeness). I suspect that a direct summand of a quasi-free module is not necessarily quasi-free. The intuitive decomposition of a submodule $W \subset V$ given by $W_k:= V_k \cap W$ fails to span all $W$ since $\oplus, \cap$ does not behave nicely (I have in mind the classical counterexample of $W=\{y =x\}$ being intersected with $V_1 = \{y=0\}$ and $V_2 = \{x=0\}$). If some 'straightening' is possible for $n=p$, it will be even harder if we leave the vector space world.