Hint: Recall the theorem highlighted below, and note that it follows that $$\quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 6} \quad \cong \quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 2}\times \mathbb Z_{\large 3}$$
This might help to make your task a bit more clear, noting that each of $\mathbb Z_2, \; \mathbb Z_3,$ and $\,\mathbb Z_6 \cong \mathbb Z_2 \times \mathbb Z_3$ are cyclic, but $\;\mathbb Z_2 \times \mathbb Z_2,\;$ of order $\,4,\,$ is not cyclic. Indeed, there is one and only one group of order $4$, isomorphic to $\mathbb Z_2\times \mathbb Z_2$, i.e., the Klein $4$-group.
Theorem: $\;\mathbb Z_{\large mn}\;$ is cyclic and $$\mathbb Z_{\large mn} \cong \mathbb Z_{\large m} \times \mathbb Z_{\large n}$$
if and only if $\;\;\gcd(m, n) = 1.$
This is how we know that $\mathbb Z_6 = \mathbb Z_{2\times 3} \cong \mathbb Z_2\times \mathbb Z_3$ is cyclic, since $\gcd(2, 3) = 1.\;$
It's also why $\,\mathbb Z_2\times \mathbb Z_2 \not\cong \mathbb Z_4,\;$ and hence, is not cyclic, since $\gcd(2, 2) = 2 \neq 1$.
Good-to-know Corollary/Generalization:
The direct product $\;\displaystyle \prod_{i = 1}^n \mathbb Z_{\large m_i}\;$ is cyclic and
$$\prod_{i = 1}^n \mathbb Z_{\large m_i}\quad \cong\quad \mathbb Z_{\large m_1m_2\ldots m_n}$$ if and only if the integers $m_i\,$ for $\,1 \leq i \leq n\,$ are pairwise relatively prime, that is, if and only if for any two $\,m_i, m_j,\;i\neq j,\;\gcd(m_i, m_j)=1$.
Suppose this non-cyclic subgroup is generated by elements (a,0), (b,1) (in other cases it would be cyclic, also any non-cyclic subgroup of a two-generated abelian group will be two-generated) and $a$ is nonzero. If $a$ generates $\mathbb{Z}_4$, then $\langle(a, 0), (b, 1)\rangle = \mathbb{Z}_4 \times \mathbb{Z}_2$. If $b$ generates $\mathbb{Z}_4$ and $a$ does not, then $\langle(a, 0), (b, 1)\rangle$ is cyclic. All other situations can be listed further:
$\langle(2,0), (0, 1)\rangle$ and $\langle(2,0), (0, 1)\rangle$ both generate the same subgroup isomorphic to Klein four-group (which appears to be the only non-cyclic proper subgroup)
Best Answer
You are on the right track. An element $(a,b) \in \mathbb{Z}_{p^r} \times \mathbb{Z}_{p^r}$ is of order $p^r$ if and only if at least one of $a,b$ is of order $p^r$ in $\mathbb{Z}_{p^r}$. Since there are $\phi(p^r)$ such elements in $\mathbb{Z}_{p^r}$, by inclusion and exclusion, we get $$2\phi(p^r)p^r-\left(\phi(p^r)\right)^2$$ such $(a,b)$.
Now dividing by $\phi(p^r)$ as you suggest, will give the answer.
EDIT
When we count the elements, the expression $$2\phi(p^r)p^r$$ counts the elements where both $a$ and $b$ are of order $\phi(p^r)$ twice, so we have to subtract them.$$|A\cup B|=|A|+|B|-|A\cap B|$$