There are a lot of misunderstanding in your question, I will try to go over them to explain them better.
The power set of $\mathbb{R}$ is often described as being "bigger" or $\aleph_2$
$\aleph_0$ is how we denote the cardinality of $ℕ$, the countable infinity.
$\aleph_1$ is the minimal well orderable cardinal that is strictly bigger than $\aleph_0$ (I will later explain what "bigger" exactly means)
$\aleph_2$ is the minimal well orderable cardinal that is strictly bigger than $\aleph_1$.
We cannot, from the usual axioms of set theory, deduct completely the relation between the cardinality of the powerset of $ℝ$ to that of $\aleph_2$, the most we can say is that $|\mathcal{P}(ℝ)|≥\aleph_2$.
It is entirely possible that $|\mathcal{P}(ℝ)|>\aleph_2$, it is also possible that $|\mathcal{P}(ℝ)|=\aleph_2$. But you are correct by saying $|\mathcal P(ℝ)|>|ℝ|$, this relation is true for every set, not only $ℝ$.
By "$a$ is bigger than $b$" ($a>b$) we mean that there is injective from $a$ to $b$, but there is no bijection between them. (while equals means that there exists a bijection).
One, what practical spaces of objects have a cardinality of $\aleph_2$?
This one is a bit tricky to answer, because there are 2 ways to look at it:
what kind of spaces have cardinality that is provable $\aleph_2$
Or
what kind of spaces have cardinality that can be $\aleph_2$
For the first variation, there are a lot of spaces (for example, we can show that there is a group of that cardinality), but most likely you won't find any example that is interesting by itself.
For the second variation: it is possible that $|ℝ|=\aleph_2$! It is also possible that the function space $ℝ^ℝ$, has cardinality $\aleph_2$.
Are there any relevant theorems about mappings to and from sets that have a cardinality of $\aleph_2$?
Not really. Or rather, not anything special to $\aleph_2$. I kind find a lot of properties of such functions, but nothing of the form you would find in analysis most likely (for example: for any function from $\aleph_2$ to the naturals, there exists some natural $n$ whose pre-image is of cardinality $\aleph_2$)
Further, does there exist a set with a cardinality of $\aleph_2$ that can be ordered, or is ordering intrinsically impossible on a set with such cardinality?
Any set of cardinality $\aleph_2$ can be well ordered. Like I said, by definition $\aleph_2$ is well orderable, so if $X$ is of cardinality $\aleph_2$, let $f:X→\aleph_2$ be a bijection, and define order on $X$ by $x<y⇔f(x)<f(y)$.
Although, saying anything about "intrinsically" is not really possible without more information about what your mean.
In other words, why study larger infinities like $\aleph_2,$ what are these bigger infinities good for?
Like I said before, it is possible that $|ℝ|=\aleph_2$, and in fact, for each $\aleph_k$, it is possible that $|ℝ|>\aleph_k$. So maybe this is a good reason for you.
About why the powerset of $ℝ$ is interesting, tons of analysis relies on spaces of subsets of reals, e.g. topology, measure, and more.
Also, it is interesting research on it's own.
Note that apart from $\aleph_n$, there is also a definition of $\beth_n$: $\beth_0$ is exactly $\aleph_0$, $\beth_1$ is the cardinality of the powerset of $\beth_0$ (note that $\beth_1=|ℝ|$), $\beth_2$ is the cardinality of the powerset of $\beth_1$, so the powerset of the reals is of cardinality $\beth_2$.
