Why are there only 5 abelian groups of order 48? 
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?

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.
