The orders $\leq_1$ and $\leq_2$ are not exactly the same.

For example, let $a$ be an ultrafilter on a set $I$ and let $I\subset J$ where $J$ is a set of much larger cardinality, and let $b$ be the filter generated by $a$ on $J$, by giving measure $0$ to the new part $J-I$. These ultrafilters are isomorphic, in the sense that they have measure one sets with a bijection between them preserving the ultrafilter (namely, the identity function on $I$), and so $b\leq_1 a$. But we don't have $b\leq_2 a$, since there is no surjection from $I$ to $J$ on cardinality grounds, and so there is no $f$ for which $b=f[a]$.

The moral of this is that your relation $\leq_2$ is simply too strong, as is your notion of isomorphism. For isomorphisms of ultrafilters and measures generally, you want to be able to ignore any measure 0 set. Thus, adding a huge cardinality collection of points to the domain, but with zero total measure, should give you an isomorphic measure, but it doesn't with your definitions. To fix this, what you want to do is to be allowed to restrict to measure one sets on both sides where you achieve the bijection and equality, and in this case, the corresponding two relations are the same. 

In the case of countably complete ultrafilters, when the Rudin-Kiesler order is prominently studied, this amounts to the classical observation, due originally to Solovay I believe, that the equivalence relation underlying the Rudin-Kiesler order, namely $\mu\leq_{RK}\nu$ and $\nu\leq_{RK}\mu$, is equivalent to the assertion that $\mu$ and $\nu$ are isomorphic by a bijection on suitably chosen measure one sets for each ultrafilter. This proof requires AC. But this observation will answer a number of your other questions for this measure context.

To prove the classical observation, it suffices to show, by composing the two functions, that whenever $\mu$ is a countably complete ultrafilter on $I$ and $f:I\to I$ for which $X\in\mu\leftrightarrow f^{-1}(X)\in\mu$, then $f$ is the identity on a set in $\mu$. To see this, view $f$ as the arrows of a directed graph on vertex set $I$. Let $X$ select one vertex from each connected component. Thus, every point in $I$ is reachable from some finite $i$ forward applications of $f$ and $j$ reverse applications of $f$ from the set $X$. Thus, by countable additivity, there must be fixed $i$ and $j$ for which that set has $\mu$-measure one, and so $X$ has $\mu$-measure one. But $f[X]$ also has $\mu$-measure one, and so $X\cap f[X]\in\mu$. From this, since $X$ contains at most one point from each connected component, it follows that $f$ is the identity on most points of $X$. 

(I'm less sure now what happens when the measures are not countably complete.)