As the Wikipedia page notes, there are two ways to approach cardinality. One, which is what you are getting at here, is to construct the cardinal numbers and have a procedure that assigns each set $S$ a unique cardinal $Card(S)$. This construction is somewhat involved and is mostly the domain of set theorists and the like. Most ordinary mathematicians think of cardinality through the relations "$A$ has the same cardinality as $B$" denoted by $|A| = |B|$ and "$A$ has cardinality less than or equal to $B$", denoted by $|A| \leq |B|$.
We define $|A| = |B|$ as $\exists \phi:A \to B, \phi$ is a bijection. And we define $|A| \leq |B|$ by $\exists \phi: A \to B, \phi$ is an injection.
Then the Schroder-Bernstein theorem gives that $|A| \leq |B|$ and $|B| \leq |A| \implies |A| = |B|$.
Now if we just consider finite sets, we can alternatively define a "function" (note it won't be a true set function, since there is no set of all finite sets) $Card(S)$ that assigns a finite set $S$ a unique natural number that is its number of elements. We then can note that $Card(A) = Card(B) \iff |A| = |B|$ and $Card(A) \leq Card(B) \iff |A| \leq |B|$, so these two approaches are the same for finite sets.
Edit: Defining the $Card$ "function" for finite sets. Since $Card$ cannot be a set function as noted above, we are really looking for a predicate $\Phi(A,n)$ s.t. $A$ is finite implies $\exists! n \in \mathbb{N}, \Phi(A,n)$.
Denote $set(n) = \{0,...,n-1\}$. Define $\Phi(A,n) \iff n \in \mathbb{N} \land \exists \phi : A \to set(n), \phi$ is a bijection.
Then to show $\Phi$ has the properties we want, we note that uniqueness comes directly from the nonexistence of bijections between $set(n)$ and $set(m)$ for $n \neq m$ and since $A$ is finite can be defined to mean $\exists n \in \mathbb{N} \exists \phi : A \to set(n), \phi$ is a bijection, we get existence of some $n$ s.t. $\Phi(A,n)$ provided $A$ is finite.
Thus $\Phi$ defines a function, since to each finite $A$, we get a unique $n \in \mathbb{N}$.
Yes, no, and consequently also no.
The first one is actually a theorem: if there is an infinite Dedekind-finite set, then there is one which can be mapped onto a strictly larger set. That means that the surjection defines the partition you are looking for.
Suppose that $A$ is Dedekind-finite, then $S_{inj}(A)$, which is the set of all injective finite sequences from $A$, is also Dedekind-finite. For if it wasn't, then we would have a countably infinite set of enumerated finite sets, and their union is therefore a countably infinite subset of $A$, and that is impossible.
Now take, for example $S_{inj}(A)$ without the empty sequence. Then the function erasing the first coordinate from each non-empty sequence is surjective onto $S_{inj}(A)$. Indeed, taking any subset which contains "all sequences of length $n\in I$" for some fixed infinite $I\subseteq\omega$ will work in a similar way.
For the second question, note that if $A$ is amorphous and $f\colon A\to X$ is surjective, then $X$ must be amorphous (or finite, if you'd like to require that amorphous implies infinite). For otherwise, simply partition $X$ into two infinite subsets and look at their preimages, both would have to be infinite subsets of $A$ which are disjoint.
How does that help us? Well, if $X$ is infinite, then the preimage of each point is finite. Now, starting with some $x\in X\setminus A$, consider $f^{-1}(x)=A_0$, then $f^{-1}(A_0)=A_1$ is also finite, and it must be distinct from $A_0$, and then we can continue by recursion and define a countably family of finite subsets of $A$. But this is impossible as the power set of an amorphous set is itself Dedekind-finite.
Note, however, that if $X$ is amorphous, its partitions are either finite; almost all singletons; or incomparable with $X$ itself. So we can get a partition that is incomparable, just not a partition that is strictly bigger.
Best Answer
If $A$ can be written as an image of $\omega$ then $A$ is either finite or countably infinite. If we assume that it is infinite then it has to be Dedekind-infinite.
The reason is that surjections from well-ordered sets can be inversed.
Note also that $A=\{n\in\omega\mid\varphi(n)\}$ is a subset of $\omega$ and therefore can only be Dedekind-finite if it is finite.
Let us make some clarifications too, while we're at it:
If you show that $A\subseteq\omega$ is bounded then it is finite. If you show that it is not Dedekind-infinite then it is finite, and bounded. If you have shown that there is some $k<m$ such that $k\notin A$ and $m\in A$ then $A$ is not an ordinal itself.