= Solution
First account for the unlabelled coefficient-sequence construction. The two <short exact sequences> of <abelian groups> are
$$
0\longrightarrow\mathbb Z\xrightarrow{\,n\,}\mathbb Z\xrightarrow{\widehat\alpha}\mathbb Z/n\longrightarrow0,
\qquad
0\longrightarrow\mathbb Z/n\xrightarrow{\,j\,}\mathbb Z/n^2\xrightarrow{\alpha}\mathbb Z/n\longrightarrow0,
\quad j([a])=[na].
$$
The <singular chain groups> of $X$ are free abelian. Applying $\operatorname{Hom}(C_q(X),-)$ therefore preserves these exact sequences, degree by degree, giving <short exact sequences of cochain complexes>. The associated <long exact sequence from a coefficient sequence> gives the displayed maps in <cohomology>; the connecting maps are the <integral Bockstein homomorphism> $\widehat\beta$ and the modulo-$n$ <Bockstein homomorphism> $\beta$. The first omitted map is multiplication by $n$, and the second is induced by $j$.
For the requested example, attach an $(i+1)$-cell to $S^i$ using a map of degree $n$. The resulting <Moore space> $X=M(\mathbb Z/n,i)$ has positive-degree <cellular chain complex>
$$
0\longrightarrow\mathbb Z\xrightarrow{\,n\,}\mathbb Z\longrightarrow0
$$
in degrees $i+1,i$. This construction also works for $i=1$, using the degree-$n$ map of the circle. In <cellular cohomology> with coefficients $\mathbb Z/n$, the differential is zero, so both $H^i$ and $H^{i+1}$ are $\mathbb Z/n$.
Lift the cochain taking value $1$ on the $i$-cell to a cochain with coefficients $\mathbb Z/n^2$. Its coboundary takes value $n$ on the $(i+1)$-cell, which is $j(1)$. The definition of the <connecting homomorphism> therefore sends the degree-$i$ generator to the degree-$(i+1)$ generator. Hence
$$
\boxed{\beta:H^i(M(\mathbb Z/n,i);\mathbb Z/n)\xrightarrow{\ \cong\ }H^{i+1}(M(\mathbb Z/n,i);\mathbb Z/n).}
$$
It is nonzero for every $i\geq1$ and $n\geq2$, including composite $n$. This is the <Bockstein on a cyclic Moore space>.
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