How to resolve Skolem's Paradox by realizing what can be said of a set is relative to what is in the domain of some model? I apologize in advanced if I'm hopelessly confused...
Skolem's Paradox, I suppose, can be put like this:
$M$ is a countable model of ZFC and $M$ implies the existence of uncountable sets.
I suppose that people find this initially paradoxical because they assume the statement is said from a single, absolute perspective.  However, there are (necessarily) two perspectives involved: the inside perspective on $M$ and the outside perspective on $M$.  Once these perspectives are separated, we realize that there is no paradox.  Consider:
The former conjunct "$M$ is a countable model of ZFC" is necessarily said from an outside perspective on $M$ -- as discussed here.  Actually, $M$ can't express its own cardinality at all.  
(Is $M$'s inability to express its own cardinality related to $M$'s being a proper class -- namely, that there is no function in $M$ that takes one of $M$'s members onto the universe of $M$?)
Continuing on…Let $N$ be the outside perspective on $M$ such that there is a bijection $f\in N$ between the domain M of $M$ and $\omega^N\in N$.
The latter conjunct "$M$ implies the existence of uncountable sets" is obviously said from $M$'s inside perspective -- after all, $M$ is a model of ZFC and thus must satisfy Cantor's Theorem.
So, we can separate the perspectives of the paradoxical statement above here:


*

*From $M$'s perspective, $M$ is a proper class and there is some $A\in M$ such that $A$ is uncountable in $M$.

*From $N$'s perspective, $M$ is countable.


These two statements are jointly consistent when we realize that what can be said of a set $B$ is relative to what some model has to say about $B$.  And so, the paradoxical statement isn't so paradoxical.
Is there anything wrong with what I have written above? (It took me a long time to learn this stuff, esp. with zero background in set-theory, higher-math, higher-logics, model-theory, etc., so specifically telling me where I am going wrong, if I am, will be a great help and a great relief.)

What I'm really interested in is what can we say about $A$ from $N$'s perspective?  Are the following possible:


*

*$M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as finite.

*$M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as countable.

*$M$ sees $M$ as a proper class, $M$ sees $A$ as uncountable, $N$ sees $M$ as countable, and $N$ sees $A$ as uncountable.


Under what conditions might (1) - (3) be individually possible (obviously they can't be jointly possible)?
I suspect this question might be rather simple.  For example, (2) could be possible when $N$ recognizes a bijection both between $M$ and $\omega^N$ and between $A$ and $\omega^N$.  (3) could be possible when $N$ recognizes a bijection between $M$ and $\omega^N$ but doesn't recognize a bijection between $A$ and $\omega^N$.
My goal here is to understand/stress the fact that truth in model theory is relative to what particular models have to say about their members.  So, I'm trying to see that while $M$ may take $A$ to be uncountable, $N$ can take $A$ to be of any cardinality even under the condition that $N$ sees $M$ as countable.
 A: The second and third are easily describable:
Suppose that $N$ is a model of ZFC and $N$ thinks that $M$ is a countable transitive model of ZFC ($M$ may not be such model, but internally to $N$ this assertion holds).
This means that $N$ thinks that $M$ is countable, and that every element of $M$ is countable. $M$, on the other hand, knows some sets which are uncountable to it. So we have $\omega_1^M$ is a countable ordinal in $N$, so the second situation holds.
Suppose now that we have a nice model of ZFC, $N$, which is uncountable and it knows about uncountable sets. If we take $\omega_1^N$ we can consider $M$ a countable elementary submodel of $N$ such that $\omega_1^N\in M$. By elementarity $M$ and $N$ agree on $\omega$ and both agree that there is no bijection between that set and $\omega_1^N$. So we have the third situation in which both models agree that some set is uncountable.
Lastly to address the first situation, what we want is to have an ill-founded model of ZFC which thinks of an uncountable set as its $\omega$, but that it will also know about some model which is nicer. I am not sure how to address this situation since for $N$ to think that $M$ is a model of ZFC, $N$ would have to assert that $M$ have certain properties in $N$. These properties may make it impossible to make the jump from infinite to finite in this manner.
There is a possibility that $N$ will not be aware that $M$ is a model of ZFC, but that defeats the purpose because then we talk from an external point of view about both these models.
A: (CAVEAT LECTOR: this post was made with the implicit assumption that $M$ and $N$ are transitive models of ZFC; that makes several of the thoughts here inaccurate.  See the comments below.)
The only possible situation here is (2).  (1) is impossible because finiteness is absolute - a set that's finite is 'always' finite. (Whoops - see below!) (3) is also impossible: since $A\in M$, and since $N$ 'sees' $M$ as a countable transitive set, $N$ also has $A\subset M$; applying the bijection between $M$ and $\omega$ to $A$ then gives a bijection in $N$ from $A$ to a countable set.
As for the correctness of what you said, I think it's worth putting more emphasis on what countability and uncountability mean in models; I would speak more explicitly about the lack of any bijection $g$ from $A$ to $\omega$ in $M$ and how $N$ can have such a bijection that $M$ is 'unaware' of.
EDIT: as pichael points out in the comments, finiteness is not absolute; being a specific finite number is, and I believe being a finite ordinal ought to be, but just the claim 'A is finite' isn't absolute across models.  On the other hand, for the specific case of $A = \left(2^\omega\right)^M$, it's still possible to show that $N$ can't think it's finite, because $M$'s inclusion of $\omega\subset A$ carries through into $N$, and $\omega$ is absolute.
