It is natural to suspect 1) to be true since it is true when "for principal ideals" is replaced with "for all ideals." This is the same as saying "if $R$ has it, then its quotient rings have it."
Since $S\cong R/I$ for some ideal $I$, we would be considering an ascending chain of principal ideals in $R/I$. For each such ideal $(x+I)\lhd R/I$, the ideal $(x)\lhd R$ maps onto $(x+I)$, and this corresponds to and ascending chain of ideals in $R$. Work to show that if that chain stabilizes, then so does the one in the quotient ring.
You're right about 2, but the example you gave doesn't show this (we'll talk about that in the third section.)
An easy way to try to make a $S$ and ring $R$ such that $R\to S$ is onto and $S$ has some property that $R$ doesn't is to make $S$ a summand of $R$ and project onto $S$. Pick a ring $T$ that doesn't have ACCP, and a ring $S$ that does have ACCP. Let $R=S\times T$ and look at the projection $(s,t)\mapsto s$ from $R\to S$.
In your first example, it looks like you are misusing some logic:
"In a UFD, if irreducible elements are prime, then it satisfies the ACCP." in the second sentence and
"If there is an irreducible element that's not prime in $S$, $S$ doesn't satisfy the ACCP." in the third sentence.
Both of these seem to be warped applications of this correct theorem:
"A commutative domain is a UFD iff it has ACCP and all irreducible elements are prime."
You are apparently thinking there is some causal relationship between the ACC condition and irreducible-implies-prime conditions, but that's not what's going on. Rather, they are two halves of conditions that make a UFD. You can have ACCP without irreducibles being prime, and you can have irreducibles all prime without ACCP.
Now for the second argument. The claim "$R$ isn't a field, so it doesn't satisfy ACCP" is just false. You probably know that the integers satisfy the ACC on all ideals, and moreover that it is a principal ideal domain. But you see that ACC on all ideals and all ideals principal implies that the integers have ACCP right?
So the projection $\Bbb Z\to \Bbb Z/2\Bbb Z$ is not a counterexample to #2. Both rings have ACCP.