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 measures are isomorphic, in the sense that they have measure one sets with a bijection between them preserving the meausre (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 we don't have $b=f[a]$.
The moral of this is that your relation $\leq_2$ is simply too strong, and what you want to do is to restrict to measure one sets on both sides where you achieve the equality, and then the relations are the same. That is, the observation is 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, and is due originally to Solovay, I believe. But this observation will answer a number of your other questions. (I'll try to post a proof of it later...)