If a filter $X$ contains any set $A$, then it contains the principal filter of $A$. Thus you are really asking: for which subsets $A$ of a set $X$ can a free ultrafilter contain $A$?
It is a standard exercise to see that such free ultrafilters exist iff $A$ is infinite.
Let me briefly sketch the proof:
If an ultrafilter contains a finite union of sets $A_1,\ldots,A_n$, then it contains $A_i$ for at least one $i$. Thus an ultrafilter which contains any finite set is principal.
If $A$ is infinite, consider the family $F$ of subsets which either contain $A$ or have finite complement. Then $F$ satisfies the finite intersection condition, so is a subbase for a filter $\mathcal{F}$ (see e.g. Exercise 5.2.5 of http://alpha.math.uga.edu/~pete/convergence.pdf). Then it can be extended to an ultrafilter which, since it contains the Frechet filter, is free.
I understand your question better now.
First, in your general context of filters the relations
$\leq_1$ and $\leq_2$ are not the same. To see this, let
$G=\{I\}$ be the trivial filter on a set $I$ with at
least two points, and let $\mu$ be any nonprincipal
ultrafilter on $I$. Since $G\subset \mu$, we see that
$\mu\leq_1 G$ as witnessed by the identity function $i$ on $I$. (Details: since $i[I]=I$, it follows that $i[G]$ is the filter with base $\{I\}$, which is the same as $G$. So $i[G]=G$, which is a subset of $\mu$, and so $\mu\leq_1 G$.) Meanwhile, I claim that $\mu\not\leq_2 G$. To see this, observe that for any function $f:I\to I$, we have $f[G]$ is the filter with base $\{f[I]\}$, and so $f[G]\neq\mu$ since $\mu$ is nonprincipal.
So the relations are
different.
Note also that if $\mu$ is an ultrafilter on $I$ and $F\leq_1 \mu$ via the function $f$ for a filter $F$, then $F$ is an ultrafilter. The reason is that if $Y\notin F$, then $f^{-1}Y\notin\mu$ and so $f^{-1}(I-Y)\in\mu$, which implies $f[f^{-1}(I-Y)]\in F$, which implies $I-Y\in F$, so $F$ is an ultrafilter.
Next, I claim that for ultrafilters, the relations are the
same.
Theorem. If $\nu$ is an ultrafilter, then $F\leq_1\nu\iff F\leq_2\nu$.
Proof. It suffices to prove the forward direction. Suppose $\nu$ is an ultrafilter on a set $J$ and $F$ is a filter on $I$ and $F\leq_1\nu$ as witnessed by $f:J\to I$. So $f[\nu]\subset F$. Consider any $X\in F$. If $f^{-1}X\in\nu$, then we get $X\supset f[f^{-1}X]\in f[\nu]$ and so $X\in f[\nu]$. Otherwise, since $\nu$ is an ultrafilter, we have $f^{-1}(I-X)\in\nu$ and so $I-X\supset f[f^{-1}(I-X)]\in f[\nu]\subset F$, which would put disjoint sets in $F$, a contradiction. QED
Finally, I claim that for ultrafilters, the relation $\leq_2$ is the same as the
Rudin-Keisler order. The usual definition of this order is
that if $F$ is a filter on $J$ and $f:J\to I$ is any
function, then one we may define a filter $G=f*F$ on $I$ by
$X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Keisler
order is defined so that $G\leq_{RK} F$ if and only if
there is $f$ for which $G=f*F$.
Suppose $F$ is a filter on $J$ and $f:J\to I$. I claim
generally that $f*F=f[F]$. This is because $Y\subset
f^{-1}f[Y]$ for $Y\subset J$ shows that $f[F]\subset f*F$;
and conversely $f[f^{-1}X]\subset X$ for $X\subset I$ shows
$f*F\subset f[F]$.
It follows that $\leq_2$ is the same as the Rudin-Keisler
order.
Best Answer
The product $a\times b$ of two ultrafilters is an ultrafilter if and only if, for every function $f$ from the underlying set of $a$ into $b$ (that's not a typo for "into the underlying set of $b$"), there is a set $A\in a$ such that $\bigcap_{x\in A}f(x)\in b$. One way for this to happen is for the underlying set of $a$ to be small enough and $b$ to be complete enough, as in Joel's answer. Notice, though, that the condition is, despite its appearance, symmetrical between $a$ and $b$. In particular, if $b$ lives on $\omega$ while $a$ is countably complete, then the condition is satisfied because $f$ will be constant on some set in $a$ (because countably complete ultrafilters are closed under intersection of continuum many sets). [Archaeologists may be interested to know that this characterization of ultrafilters whose product is ultra occurs on page 22 of my 1970 Ph.D. thesis, which is, thanks to patient scanning, available from my web page.]