Intuition behind Principal Ideal Domain In a group, normal subgroups are most likely not cyclic and hence not generated by a single element. Ideals, as kernels of homomorphisms, is a similar concept to normal subgroups. Why would we expect some integral domains to have the property which all ideals are principle (generated by one single element)?
As a second part of the question, what is the best way to visualize principle ideal domains in general? Since prime ideals are maximal, these ideals "cover a lot of grounds" and their intersections contain the reducible elements? I do not know whether my visualization is correct.
 A: Let's break this down. The first question that you seem to have is why would we expect some integral domains to be principal. This seems like a strong thing to hope for because it is. However, if you have learned what a Euclidean domain is (this is a ring that has a division algorithm), it can be shown that all Euclidean domains are PIDs, so this gives us that $\mathbb{Z}$ is a PID, and if you recall polynomials over fields have a division algorithm, so if $F$ is a field $F[x]$ will be a PID, so polynomial rings (in a single variable) will end up being principal.
The reasons why we might care about PIDs is that there is a theorem that all PIDs are unique factorization domains, so there is unique factorization in PIDs. If you study algebraic number theory there is this notion of a class group which is defined as fractional ideals modulo principal ideals, and in some sense this group measures how far away the ring of integers of a number field is from having unique factorization.
To answer the last part of your question on visualizing principal ideals. I'm not entirely sure what you mean in terms of visualizing, but I think one way to view this is through the lens of the spectrum of the ring (which goes towards the realm of algebraic geometry). If $R$ is a ring, then $\text{Spec}(R)=\{\mathfrak{p}\subset R\}$ that is the spectrum is the set of prime ideals of the ring. However, we don't just consider Spec as a set, we give it a topological structure whose closed sets are $V(\mathfrak{a})=\{\mathfrak{p}\supset\mathfrak{a}\}$ where  $\mathfrak{a}\subset R$ is some arbitrary ideal. If you think about it the maximal ideals correspond to closed points in $\text{Spec}(R)$, and since a PID is one where every (non-zero) prime ideal is maximal every point is a closed point.
An example of this would be given the ring $\mathbb{C}[x]$, then we have that this is a PID, and there is this theorem known as Hilbert's Nullstullensatz that says that the maximal ideals of $\mathbb{C}[x]$ are exactly those of the form $(x-a)$ where $a\in \mathbb{C}$. Thus, if we look at $\text{Spec}(\mathbb{C}[x])$, the points correspond to just the points of $\mathbb{C}$ (and the zero ideal is a "fuzzy" point as all prime ideals contain it).
I hope this helps and gives you a bit of intuition in your studies.
