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 $\,m_i, m_j,\;i\neq j,\;\gcd(m_i, m_j)=1$.
Yes, what you did is correct, but you missed some subgroups.
The idea is to note that the subgroups you are looking for are of the form:
\begin{equation}
C_x\times C_y\times C_z
\end{equation}
and you want
\begin{equation}
C_x\times C_y\times C_z \cong C_{30}
\end{equation}
with $x\in\{1,2,3,6\}$, $y\in\{1,2,5,10\}$ and $z\in\{1,3,5,15\}$. You can take $x,y,z$ in the set of all respective divisors because the groups $\mathbb Z_6$, $\mathbb Z_{10}$, $\mathbb Z_{15}$ are cyclic.
Thanks to the Chinese remainder theorem the problem is equivalent to find triple $(x,y,z)$ (taken in the sets above) such that $xyz=30$. By a direct computation we get $8$ triples:
\begin{gather}
(1,2,15)\\
(1,10,3)\\
(2,5,3)\\
(2,1,15)\\
(3,10,1)\\
(3,2,5)\\
(6,5,1)\\
(6,1,5)
\end{gather}
That correspond to all the subgroups you are looking for.
The number of cyclic subgroups of order $30$ in $G$ is bigger. To calculate this number is sufficient to count all the element of order $30$ in $G$ and divide this number by $\varphi(30)$ because every cyclic subgroup of order $30$ has exactly $\varphi(30)$ generators.
Since $G\cong \mathbb Z_2\times\mathbb Z_2\times\mathbb Z_3\times\mathbb Z_3\times\mathbb Z_5\times\mathbb Z_5$ the number of element of order $30$ is
$$
(2^2-1)(3^2-1)(5^2-1) = 3\cdot 8\cdot 24
$$
So the number of cyclic subgroups is:
$$
\frac{3\cdot 8\cdot 24}{\varphi(30)}= \frac{3\cdot 8\cdot 24}{1\cdot 2\cdot 4}=72
$$
Best Answer
The number of non-cyclic such subgroups is $0$: an abelian group of order $14$ is cyclic (by the structure theorem, CRT or some other simple arguments).
An element $(a,b)$ has order $\rm{lcm}(|a|,|b|)$. So we get the possibilities $(2,7),(7,2),(2,14),(14,2),(7,14),(14,7),(1,14),(14,1)$ and $(14,14)$. Note that in a cyclic group there are $\varphi(2)=1, \varphi(7)=6$ and $\varphi(14)=6$ elements of order $2,7$ and $14$ respectively.
Thus we get $6+6+6+6+36+36+6+6+36=144$. But to finish we need to divide by $\varphi(14)=6$, since there are that many elements of order $14$ in each such subgroup. Thus we get $24$.