Point 1. What you are asking is not what you claim to be trying to understand.
Point 2. $\mathbb{Z}_3\oplus\mathbb{Z}_5$ is isomorphic to $\mathbb{Z}_{15}$. What makes you say it isn't?
Point 3. Perhaps you want to compare with $\mathbb{Z}_3\oplus\mathbb{Z}_6$, which is not isomorphic to $\mathbb{Z}_{3\times 6}=\mathbb{Z}_{18}$.
The reason is that while $2$ and $3$ are relatively prime, $3$ and $6$ are not. Every element of $\mathbb{Z}_a\oplus\mathbb{Z}_b$ has order that divides $\mathrm{lcm}(a,b)$. If $\mathrm{lcm}(a,b)\lt ab$, then the group cannot be cyclic, because a finite group of order $k$ is cyclic if and only if there is (at least) one element of order exactly $k$. If every element has order dividing (and hence, less than or equal) to something strictly smaller than the order of the group, then the group cannot be cyclic.
Since $\mathrm{lcm}(a,b)=ab$ if and only if $a$ and $b$ are relatively prime, this tells you that a necessary condition for $\mathbb{Z}_a\oplus\mathbb{Z}_b$ to be cyclic is for $\gcd(a,b)=1$. In order to show that it is enough (sufficient), then notice that the order of $(1,1)$ is exactly $\mathrm{lcm}(a,b)$:
$$\begin{align*}
k(1,1) = (0,0)\text{ in }\mathbb{Z}_a\oplus\mathbb{Z}_b&\iff (k,k)=(0,0)\text{ in }\mathbb{Z}_a\oplus\mathbb{Z}_b\\
&\iff k\equiv 0\pmod{a}\text{ and }k\equiv 0\pmod{b}\\
&\iff a\text{ divides }k\text{ and }b\text{ divides }k\\
&\iff \mathrm{lcm}(a,b)\text{ divides }k.
\end{align*}$$
So the order of $(1,1)$ is exactly $\mathrm{lcm}(a,b)$. Hence, if $\mathrm{lcm}(a,b)=ab$, then $\mathbb{Z}_a\oplus\mathbb{Z}_b$ has (at least) one element of order $ab$, and so it is cyclic.
As a consequence, we have that in (finite) direct sum of cyclic groups,
$$\mathbb{Z}_{a_1}\oplus\cdots\mathbb{Z}_{a_k}$$
is cyclic, and isomorphic to $$\mathbb{Z}_{a_1\times\cdots\times a_k}$$
if and only if the $a_i$ are pairwise relatively prime.
Point 4. The question you say you are trying to understand (When two groups of the form $\mathbb{Z}_n$ are isomophic) is: $\mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_m$ if and only if $n=m$. They have to have the same size to be isomorphic; and if they are the same size and both cyclic, then they are isomorphic.