I am trying to understand exactly how this works. I am following this procedure up until it shows the second equivalences. For example, why is $\mathbb{Z}_{2^3}\times \mathbb{Z}_{2} \times \mathbb{Z}_{3} \cong \mathbb{Z}_2 \times \mathbb{Z}_{24}$? Furthermore, why isn't it isomorphic to $\mathbb{Z}_{48}$, or any of the other products on the RHS of the second equivalence?
$48=2^4×3$. The number of partitions of $4$ is $5$. They're $$1+1+1+1, 1+1+2,2+2,1+3,4$$.
Now apply the structure theorem. It implies (along with the Chinese remainder theorem) the general result that if $n$ has prime factorization $p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, then the number of abelian groups of order $n$ is $p(a_1)p(a_2)\cdots p(a_k)$, where $p(x)$ is the number of partitions of $x$.
I have to assume that "Theorem 51" is the classification of finite abelian groups. If we recall, this says that
Every finite abelian group can be written as a direct sum of cyclic groups of prime power order
$$ \bigoplus_\alpha \mathbb{Z} \big / p_\alpha^{k_\alpha} \mathbb{Z}$$
Moreover, this decomposition is unique up to (noncanonical) isomorphism
So, let's start small. If we want to understand all abelian groups of size $12$, we can factor $12 = 2^2 \cdot 3$. Now there are two ways to get all of these prime powers:
The "uniqueness" clause of the classification tells us that these decompositions are not isomorphic. Of course, we can also see this directly: the latter has an element of order $4$ while the former doesn't. So they cannot be isomorphic!
Let's see another example. What are the possible abelian groups of order $2^3 \cdot 3^2 \cdot 7$?
Well, we have to have $3$ copies of $\mathbb{Z}/2\mathbb{Z}$ somehow. We can do this as either
(do you see why?)
Then we have to have $2$ copies of $\mathbb{Z} / 3 \mathbb{Z}$. We can do this is either
Lastly, we have to have $1$ copy of $\mathbb{Z} / 7 \mathbb{Z}$, and there's only one way to do this
All in all, this gives us $6$ groups of this order (again, do you see why?)
In general, then, how do we figure out all the possible abelian groups of order $n$?
As for your followup question, "why is $\mathbb{Z} / 8 \mathbb{Z} \oplus \mathbb{Z} / 3 \mathbb{Z} \cong \mathbb{Z} / 24 \mathbb{Z}$". The answer is the Chinese Remainder Theorem.
I hope this helps ^_^
The fact you need is the Chinese Remainder Theorem: if $(m,n)=1$, then $\mathbb{Z}/mn\mathbb{Z}\cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$. Here's a quick proof: we have a map $\pi: \mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ given by $\pi(a)=(\bar{a}, \bar{b})$. Clearly $mn\mathbb{Z}\subset \ker\pi$. Conversely, if $x\in\ker\pi$, then $x$ is divisible by both $m$ and $n$. Since $m$ and $n$ are coprime, $x$ must be divisible by their product $mn$. Thus $mn\mathbb{Z}=\ker\pi$. Now it suffices to show that $\pi$ is surjective. Let $(\bar{r},\bar{s})\in \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$. Since $m$ and $n$ are coprime, we can use Bezout's identity to find $p,q\in\mathbb{Z}$ so that $pm+qn=1$.Then $\pi(spm+rqn)=(\bar{r},\bar{s})$.
We can use the Chinese Remainder Theorem to establish some of these isomorphisms you're seeing. For instance $\mathbb{Z}/8\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/24\mathbb{Z}$ since 8 and 3 are coprime.
