This is an excellent and interesting question! You are asking whether the 2-step iteration of a normal measure μ on a measurable cardinal κ is uniquely factored by the steps of the iteration itself.

The answer is **Yes**.

Let me denote κ_{0} just by κ and j_{02} by j. Since
μ_{1} is a measure in M_{1}, it has the
form j_{01}(m)(κ), where m =
(ν_{α} | α < κ). Since you
have said that μ_{1} is not in
ran(j_{01}), we may choose the
ν_{α} to be all different, and different
from μ_{0}. In this case, there is a partition
of κ as the disjoint union of X_{α},
with X_{α} in ν_{α} and none
in μ_{0}. Let x = (X_{α} | α
< κ). Note that κ is not in
j_{01}(X_{α}) for any α <
κ, and similarly κ_{1} is not in
j(X_{α}). But κ is in
j_{01}(x)(β) for some β <
κ_{1}, since this is a partition of
κ_{1}. Apply j_{12} to conclude that
κ_{1} is in j(x)(β) for this β.
Thus, there is some β in the interval [κ,
κ_{1}) having the form β =
j(f)(κ_{1}) for the function f that picks the
index. From this, it follows from normality of
μ_{0} that we can write κ =
j(g)(κ_{1}) for some function g, since any
β < κ_{1} generates κ via
j_{01}. In my favored terminology, the seed
κ_{1} generates κ via j and in fact
generates all β in [κ,κ_{1}) via
j.

Similarly, suppose that δ is in the interval
[κ_{1},j(κ)). We know δ =
j_{12}(f)(κ_{1}) for some function f
on κ_{1} in M_{1}. We also know f =
j_{01}(F)(κ) for some F in V. Thus, δ =
j(F)(κ, κ_{1}). In Y, let
(α,β) be the smallest pair with δ =
j(F)(α,β). It cannot be that both are below
κ_{1}, since this would be inside
ran(j_{12}) and so the least pair must have β
= κ_{1}. Thus, δ generates
κ_{1}, which we already observed generates
κ.

To summarize, every ordinal in the interval
[κ_{1},j(κ)) generates
κ_{1}, which generates all the ordinals
β in [κ,κ_{1}), any of which
generate κ and all the other such β.

This is enough to answer your question. The k " N in your
question is just an arbitrary elementary substructure of
M_{2} containing ran(j), so suppose we have Y
elementary in M_{2} and ran(j) subset Y. The case Y
= ran(j) is one of your cases. Otherwise, Y has something
not in ran(j). Every object in M_{2} has form
j(h)(κ,κ_{1}) for some function h, so
by looking at the smallest pair of ordinals to generate a
given object with j(h), we see that there must be ordinals
below j(κ) in Y. If Y contains any ordinal δ in
the interval [κ_{1},j(κ)), then it will
contain both κ and κ_{1}, since we
observed that any such δ generates these ordinals. In
this case, Y = M_{2}, since those two ordinals
generate everything. So we assume that Y contains no such
δ. In this last case, Y must contain some ordinal
β in the interval [κ,κ_{1}). Since
any such β generates κ, Y contains all such
ordinals. It follows that ran(j_{12}) subset Y and
in fact = Y, since if Y contained anything more it would
have to have an additional ordinal δ in
[κ_{1},j(κ)).

So we've seen that your three cases are the only
possibilities. And like your previous question, there is no
need to assume that Y or N is somehow internally definable.

By the way, this was a problem that I had solved many years
ago for my dissertation, although perhaps other people had
also thought about it. I was interested in understanding
which pairs of ordinals (α,β) generate product
measures via an embedding j, and this question is very much
related to that.

(Click the edit history to see my previous answer, which was just about the case when μ_{1} is in the range of j_{01}, a case for which the answer is no.)