Quotient ring $\frac{\mathbb{Z}_n[x]}{⟨f(x)^2⟩}$

For any field $k$ and nonzero polynomial $p \in k[x]$, the quotient ring $k[x]/(p)$ is a $k$-algebra of dimension $n:=\deg(p)$. In particular, as a $k$-vector space (hence as additive group) it is isomorphic to $\underbrace{k \, \oplus ... \oplus  \,k}_{\deg(p)}$. -- Actually, if $p$ is monic, there is an "explicit" description of this $k$-algebra with matrices: Mapping $x$ to the companion matrix $C(p)$ of $p$ induces an isomorphism from $k[x]/(p)$ to the subring of $M_n(k)$ with $k$-basis $Id_n, C(p), C(p)^2, .., C(p)^{n-1}$. If you can multiply these matrices, you know everything you can know about the ring. But it's rather clear this translates the problem without necessarily making it easier: For example, nowhere in the above have we used whether $p$ is irreducible over the field $k$, which of course makes a big difference in the ring structure. This just goes to show these "explicit matrices" might not make everything about the ring "actually explicit".

So let's start over and look at the special case $f(x)=x$ first. Again for any field $k$, the ring $R:= k[x]/(x^2)$ is sometimes called the ring of "dual numbers" over $k$. We can write every element as $$a+bx$$ with unique $a,b \in k$, with obvious addition, and multiplication given by $$(a+bx)(c+dx) = ac + (bc+ad)x.$$ This is visibly equivalent to giving the underlying vector space $k \oplus k$ (cf. 1) the multiplication $$(a,b) \cdot (c,d) = (ac, bc+ad).$$ Here, the companion matrix is really easy and indeed we get to "see" this $k$-algebra as the commutative subalgebra $$\lbrace \pmatrix{a&0\\b&a}: a,b \in k \rbrace$$ of $M_2(k)$. In any case, the ring is a local $k$-algebra of dimension $2$ whose maximal ideal $\mathfrak m$ is generated by a nilpotent element whose square is zero. Both its residue field and its maximal ideal are isomorphic to $k$ as $k[x]$-modules, i.e. we have an exact sequence $$0 \rightarrow k \simeq (x)/(x^2) \xrightarrow{\subset} k[x]/(x^2) \xrightarrow{\text{mod } x} k \rightarrow 0.$$ This is an exact sequence of $k[x]/(x^2)$-modules (actually, of $k[x]$-modules) and also one of $k$-modules. Viewed as the latter, it splits by default, the splitting being induced by the natural inclusion $k \subset k[x]$ (this is just another way to say that the underlying $k$-vector space is $k \oplus k$). It does not split as sequence of $k[x]$- or $k[x]/(x^2)$-modules.

Side note for a flawed attempt in the question: The difference between $\mathbb Z/(p^2)$ and $\mathbb F_p[x]/(x^2)$. Admittedly, they have a lot in common. Both are artinian local rings whose maximal ideal is generated by an element $e$ with $e^2=0$, with residue field $\mathbb F_p$, and the maximal ideal is also isomorphic to $\mathbb F_p=R/\mathfrak m$ as module over the ring. I.e. both fit into exact sequences $$0 \rightarrow \mathbb F_p \rightarrow R_i \rightarrow \mathbb F_p \rightarrow 0$$ -- but exact sequences of what? For each $R_i$, this is an exact sequence of $R_i$-modules. But only for $R_2 = \mathbb F_p[x]/(x^2)$ is this also an exact sequence of $\mathbb F_p$-modules; in fact, $\mathbb Z/(p^2)$ (unlike $R_2$) does not have a natural structure of an $\mathbb F_p$-module at all -- if it had, each element in it would have additive order $p$. And relatedly, while the sequence of $R_2$-modules splits, the sequence of $R_1$-modules does not, not even the underlying sequence of additive groups splits: Because that would mean that there is an iso $\mathbb Z/(p^2) \simeq \mathbb Z/p \oplus \mathbb Z/p$ which is impossible because the left hand side has elements of order $p^2$, while the right hand side only has elements whose order is $1$ or $p$.

Now in general, what is $k[x]/(f^2)$ for an irreducible $f$? Let us call $F:= k[x]/(f)$ the field extension of $k$ given by that $f$. Well, a lot of the above goes through and $k[x]/(f^2)$ is a local $k$-algebra of $k$-dimension $2\deg(f)$ whose maximal ideal $\mathfrak m$ is generated by a nilpotent element whose square is zero. Both its residue field and its maximal ideal are isomorphic to $F$ as $k[x]$-modules, i.e. we have an exact sequence $$0 \rightarrow F \simeq (f)/(f^2) \xrightarrow{\subset} k[x]/(f^2) \xrightarrow{\text{mod } f} F \rightarrow 0.$$ This is an exact sequence of $k[x]/(f^2)$-modules (actually, of $k[x]$-modules) and also one of $k$-modules. However, in general we cannot view it as a sequence of $k[x]/(f) = F$-modules because the object in the middle does not have a natural structure as module over that ring. But as $k$-vector spaces, it splits by default and/or via 1 we can write the object in the middle as $$\underbrace{k \, \oplus ... \oplus  \,k}_{\deg(p)} \oplus \underbrace{k \, \oplus ... \oplus  \,k}_{\deg(p)} .$$ We want to write this just as $$k[x]/(f^2) \simeq F \oplus F$$ which we can do as long as we remember this is just an identification of $k$-vector spaces, i.e. we cannot expect the multiplication on this object to be $F$-linear. (Anyway, that being remembered, an obvious explicit such identification is to choose $\alpha \in F$ a root of $f$, then every element $d \in D$ can be written uniquely as $a_0 +a_1 \alpha + a_2 \alpha^2 + ... a_{\deg(f)-1} \alpha^{\deg(f)-1}$, and we send that element to (the residue modulo $(f^2)$ of) $a_0 +a_1 x+ a_2 x^2 + ... a_{\deg(f)-1} x^{\deg(f)-1}$.) -- We would like to endow this $$F \oplus F$$ with a multiplication $$(a,b) \cdot (c,d) \stackrel{?}= (ac, bc+ad)$$ but we cannot expect this to work. Why? There's no problem in identifications of both $k[x]/(f)$ and $(f)/(f^2)$ with $F$ as in the exact sequence above. Alternatively, one could argue with Euclidean algorithm to explicitly get elements $a,b \in k[x]$ with degree $\le \deg(f)-1$ to write an element in $k[x]$ (or $k[x]/(f^2)$) as $a +bf$ modulo $(f^2)$). But if we write those elements as polynomials and claim the calculation $(a + bf)(c+df) = ac+(ad+bc)f$ modulo $(f^2)$, we overlook that unlike in no. 2, for example $ac$ is not necessarily well-defined modulo $f^2$, only modulo $f$. In other words, one can indeed multiply $$(a,b) \cdot (0,d) = (0, ad)$$ but $(a,0) \cdot (c,0)$ is in general $(ac, m(a,c))$ where  $m(a,c)$ is a non-zero map $F \times F \rightarrow F$ which certainly depends on the given polynomial $p$ and furthermore might depend on the given identifications. This is awkward, but if taken for granted at least gives us an "explicit" multiplication $$(a,b) \cdot (c,d) = (ac, ad+bc + m(a,c)).$$ To unravel this for a given polynomial $p$ might still not be significantly easier than the approach with companion matrices in no. 1.