Assume we're in an integral domain. (I realize your example is actually a polynomial ring over ${\mathbb C}$, but let's work in a more general domain for now.) Let's also suppose our domain to be an algebra over a field of characteristic $\neq 2$.

Let's look for ideals $J$ and $K$, and an element $x$ such that $xK\subset JK$ but $x\notin J$. (This would give counterexamples to both (1) and (2).)

This is surely impossible if $K$ is principal, so let's investigate the case where $K$ is generated by 2 elements.

Then I claim the following are equivalent:

1) The ring $S$ contains a counterexample to your (1) and/or (2) with $K$ two-generated.

2) The ring $S$ contains elements $A,B,C,D,F$ with $(A-D-F)(A-D+F)=4BC$ and $F\notin (A,B,C,D)$.

${\bf Proof:}$ Let $\alpha, \beta$ generate $K$. Then given a counterexample, we can write
$$x\pmatrix{\alpha\cr\beta\cr}=\pmatrix{A&B\cr C&D\cr}\pmatrix{\alpha\cr\beta\cr}$$
for some $A,B,C,D\in J$, which we might as well assume generate $J$. Thus $x$ is an eigenvalue of the displayed two-by-two matrix and so satisfies its characteristic equation, whence there exists $F$ with $x=A-D-F$ and $(A-D-F)(A-D+F)=4BC$. Also, $(\alpha,\beta)$ must be the transpose of an eignvector, which we can take to be $(A+D-F,-C)$.

This will be a counterexample iff $x\notin J$, hence iff $F\notin J$. QED.

Thus, for algebras over a field $k$ of characteristic $\neq 2$, the universal counterexample is given by
$$R=k[A,B,C,D,F]/((A-D-F)(A-D+F)-4BC)$$
$$x=A-D-F$$
$$J=(A,B,C,D)$$
$$K=(A+D-F,-C)$$

Your ring $S$ will contain a counterexample (with $K$ two-generated) iff it contains a homomorphic image of $R$ in which $F\notin (A,B,C,D)$. When $S$ is a polynomial ring, I'm not sure whether this is the case but it might not be too hard to settle.

notmean a general mathematical context or motivation (say definitions of the involved notions) but by contrast an explanation why you would like to know this. – user9072 Mar 6 '13 at 12:39