Throughout the post, I keep to the standard assumption that UFDs, PIDs, EDs and integral domains all refer to commutative rings. (But of course, there are noncommutative domains and PIDs and even some of the others, if you work hard enough ;) )
$R[X]$ is a UFD when $R$ is
For $B$, you can find in many commutative algebra texts that a commutative ring $R$ is a UFD iff $R[x]$ is. (For example, Corollary 16.20 in Isaacs Graduate Algebra)
Why is it clear, that a principal ideal domain is a integral domain?
Look back at your definitions: a principal ideal domain is just an integral domain with an extra property (having all ideals principal). A PID is a fortiori an integral domain.
After reading what you described about your definition, it sounds like maybe this didn't make it into your notes. A principal ideal ring is a ring in which all ideals are principal, but such a ring doesn't have to be a domain (For example, $\Bbb Z/\Bbb 4$ is a principal ideal ring, but not a domain, since $2^2=0$.) A (commutative) principal ideal domain is just a (commutative) principal ideal ring that is also a domain.
Hierarchy of properties
For $C$: Making a hierarchy like this is really good exercise. (In fact, I've embarked on pictures like that with dozens of ring types.) However, I hope you're not under the impression that you are going to organize all ring types linearly.
All of the domains you mentioned are subclasses of commutative rings, but the class of division rings is not contained in commutative rings. Out of all the rings you mentioned, there is one branch containing the domains:
$\text{field}\subseteq \text{Euclidean domain}\subseteq PID\subseteq UFD\subseteq\text{domain}\subseteq \text{commutative ring}\subseteq \text{ring}$
and then there is another branch
$\text{field}\subseteq\text{division ring}\subseteq\text{ring}$
You wrote that a PID "does not have a euclidean function" which is a bit like concluding that a rectangle does not have four equal side lengths. A PID does not necessarily have a euclidean function, but it might. Just like rectangles might have four equal sides, and hence be both squares and rectangles. You should just keep in mind that a Euclidean domain has more stringent structure than a PID, since they are a special subcase. Similar comments can be made about what you wrote about a UFD not having all ideals principal, etc.
$R[X]$ not a field
For $D$: To easily see that $R$ is not a division ring, just ask yourself if you can invert $X$ or not. When you multiply polynomials together, you're only going to get higher degrees of $X$. How will you get back down to $1$?
Inheritance
People have already pointed out how it's pretty easy to prove that $R$ is a domain iff $R[x]$ is, or the same for commutativity, and for the UFD property. Just in the last section we see that the case is not so for "being a field". Someone has also given an example that whlie $F[x]$ is a PID, $F[x][y]$ is not, so that property isn't preserved either. The same is also true for Euclidean domains since $F[x]$ is actually an example of a Euclidean domain.
The ring map $A/\mathfrak{p}\rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is injective because the inverse image of $\mathfrak{p}A_\mathfrak{p}$ in $A$ (under the localization map $A\rightarrow A_\mathfrak{p}$) is equal to $\mathfrak{p}$ (one of the first things one proves about the prime ideal structure of a localization). So there is a unique $A/\mathfrak{p}$-algebra map $\mathrm{Frac}(A/\mathfrak{p})\rightarrow A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$, necessarily an injection since the source is a field. But given $a/s+\mathfrak{p}A_\mathfrak{p}$ in the target, $s\notin\mathfrak{p}$ means $s+\mathfrak{p}$ is non-zero in $A/\mathfrak{p}$, so $a+\mathfrak{p}/s+\mathfrak{p}\in\mathrm{Frac}(A/\mathfrak{p})$, and this maps to $a/s+\mathfrak{p}A_\mathfrak{p}$ under the $A_\mathfrak{p}$-algebra map obtained above, so that map is an isomorphism.
Said somewhat differently, though more or less equivalently, if $I$ is an ideal of $A$, $S$ a multiplicative set, then the natural map $S^{-1}A\rightarrow\bar{S}^{-1}(A/I)$, where $\bar{S}$ is the image of $S$ in $A/I$, is surjective with kernel $S^{-1}I$, the ideal of $S^{-1}A$ generated by $I$, and so induces an isomorphism $S^{-1}A/S^{-1}I\cong\bar{S}(A/I)$ of $A/I$-algebras. Taking $S=A-\mathfrak{p}$, and $I=\mathfrak{p}$, you get an isomorphism $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\cong(A/\mathfrak{p})_{\overline{A-\mathfrak{p}}}$. But $\overline{A-\mathfrak{p}}$ is exactly the set of non-zero elements of $A/\mathfrak{p}$, so the localization at $\overline{A-\mathfrak{p}}$ is the field of fractions of $A/\mathfrak{p}$. This isomorphism $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\cong\mathrm{Frac}(A/\mathfrak{p})$ is the inverse to the one defined above.
Best Answer
When you write $\Bbb{Z}_m[x]/f(x)$, this is the same thing as writing $\Bbb{Z}_m[x]/(f(x))$. In particular, you are taking the quotient of the ring by the ideal generated by $f(x)$. It is true in general that the quotient of a commutative ring by an ideal is again a ring. E.g. if $A$ is a ring and $\mathfrak{a}$ is an ideal, then $A/\mathfrak{a}$ is a ring. So, what you have written will always be a ring, regardless of the coefficients used. That is, for any ring $A$, $A[x]/(f(x))$ is a ring, for any $f(x)\in A[x]$.