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 $G\leq_1\mu$ as witnessed by the identity function on $I$. But for any function $f:I\to I$, it must be that $f^{-1}(j)\notin \mu$ for at least one $j$, since these are disjoint, and so $f[\mu]$ will contain $I-\{j\}$, meaning $f[\mu]\neq G$. So $G\not\leq_2\mu$, and the relations are different.
Next, I claim that for ultrafilters, the relations are the same. Suppose that $\mu\leq_1\nu$ and $\mu$ is an ultrafilter on $I$ and $\nu$ a filter on $J$. So there is a function $f:I\to J$ for which $\mu\of f[\nu]$ in your sense. Since $f[\nu]$ is a filter and $\mu$ is an ultrafilter, this implies $\mu=f[\nu]$, and so $\mu\leq_2\nu$ as desired.
Moreover, I claim that $\leq_2$ is the same as the Rudin-Kiesler order. The usual definition of this order is that if $F$ is a filter on $I$ and $f:I\to J$ is any function, then one we may define a filter $f*F$ on $J$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Kielser 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 $I$ and $f:I\to J$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset I$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset J$ shows $f*F\subset f[F]$.
It follows that $\leq_2$ is the same as the Rudin-Kiesler order.
And the $\leq_1$ order is simply a combination of the subfilter relation and the Rudin-Kiesler order.