Answer to the quick version. Yes it is true as soon as $(X,\mu)$ is a Lebesgue space, and this follows from the ergodic decomposition theorem, together with the classification of measurable partitions.
Recall that if T is an invertible measurable transformation acting on a Lebesgue space X, then there is a measurable partition $C_i$ (which may have uncountably many elements) and probability measures $\mu_i$ on $C_i$ such that all $C_i$ are invariant by T, T is ergodic w.r.t $\mu_i$ and $\mu$ is obtained by integrating the $\mu_i$. $$\mu(A) = \int_X \mu_i(A) d\mu$$ There are two kinds of ergodic components $C_i$. The one of positive measure, there are at most countably many such components.
Let us removing these components from $X$. Rohlin structure theorem on measurable partitions (1947) now says that there is a isomorphism between $([0,1]\times [0,1], \lambda)$ and $(X,\mu)$ such that the pullback of the measurable partition $(C_i)$ is the decomposition into horizontal lines $([0,1]\times \{i\})_{i\in [0,1]}$. A reference is the book of Parry, "entropy generators in ergodic theory".
The result of Rohlin is almost forgotten because you don't need it in practice. If it happens that a result is true for an ergodic transform, then it is true for an arbitrary transform in restriction to its ergodic components, and you (may) recover the result on the whole space $X$ just by using the integral formula given above.
Finally the result you are alluding in your last question is a section theorem. Given a measure preserving transform between two Lebesgue spaces X and Y, there is a section from Y to X, up to null sets, and some warning is in order here because this is not true in the Borel category. I think this is again due to Rohlin, and it can be deduced from its structure theorem for measurable partitions. Have a look at the book of Parry, but really this is overkill.