I am working on extensions in general but for sake of simplicity we can assume it's a module here. I am interested in an extension of the form $$0\to B \to E_2\to E_1 \to A \to 0$$ which is an extension of $A$ by $B$. I want to find an example of this that is relatively "simple", I tried some modulos like $\mathbb{Z}_3$ by $\mathbb{Z}_3$ but that didn't work out as I realized it doesn't have it, if I did it correctly.
Are there any good simple module examples like that to illustrate such an extension?
Here is one example. Let $R = k[x]/x^2$ and $A = B = k$ with $xA = xB = 0$. Then we have a free resolution $\cdots \to R \overset{x} \to R \overset{x}\to R \to A$, hence $Ext_R^2(A,B) = k$ generated by the quotient map $R \to k$. The corresponding extension is: $$k \overset{x}\to R \overset{x}\to R \to k.$$ Similar extensions generate the higher Exts as well.
Note: This example is nice because you can quickly calculate the algebra structure $Ext_R(k,k) \cong k[t]$.
More generally, if you take a projective resolution $$\cdots \to P_3 \overset{f}\to P_2 \to P_1 \to P_0 \to A\to 0$$ and truncate it to $$0\to B \to P_1 \to P_0 \to A\to 0$$ where $B = \text{coker}(f)$, this will give an element of $Ext^2(A,B)$ which is trivial if and only if $B \to P_1$ is a split injection. Being a split injection means $P_1/B$ is projective, so that $A$ has a projective resolution of length 1 given by $P_1/B \to P_0 \to A$, in particular $A$ has projective dimension at most 1.
If $R$ is a regular local ring, then the projective dimension of the residue field $R/\mathfrak m$ equals the Krull dimension of $R$. Thus taking a regular local ring of dimension at least 2 and using the natural Koszul resolution will give you many more examples.