I have questions (in bold) regarding the set theoretic definitions of a universe.
It's my understanding a universe is a class of the form: $$U(X) = \cup_{i=1}^\infty S_i$$ $$S_0 = X$$ $$S_{n+1} = S_n \cup \mathcal P(S_n)$$ For any assignment of $X$ this class $U(X)$ will contain (among other objects) all hereditarily finite sets (all objects are sets) which include Von Neumann ordinal numbers, ordered tuples and relations. The universe $U$ can be extended to $U_{\beta}$ so it will contain the n-ary Cartesian Product by defining $U_{\beta} = \cup_{i=1}^\infty U^{(i)}(X)$ with function composition. It's my understanding that $V_{\omega+\omega} = U(V_\omega)$ are sets usually required for normal mathematics. Why does $U(V_\omega)$ contain the sets required for usual (what is usual) mathematics? Why does this equality hold?
I also understand that $V_{\omega+1} = \mathbb N$ and $V_{\omega+2} = \mathbb R$. Are these equalities true and why do they hold?
ZFC uses the Von Neumann hierarchy defined as
$$V_0 = \varnothing$$$$ V_{\beta+1} = \mathcal{P} (V_\beta)$$$$ V_\lambda := \bigcup_{\beta < \lambda} V_\beta$$ where $\lambda$ is a limit ordinal.
$V$ is the proper class of all sets (The Universe) and $Ord$ is the proper class of ordinals. Under axioms of ZFC every set is contained in $V$ using the axiom of regularity. $$ V = \cup_{a \in Ord} V_\beta $$
It's my understanding that $V_\omega \subset U(X)$ since $V_\omega = U(\emptyset)$. Why does this equality hold?
I will give some hints and offer some corrections.
First off, to repeat my last answer to a similar question of yours, the use of the symbol $\infty$ is ambiguous and ill-defined in the context of Set Theory. From now on, I advice you to never use the symbol $\infty$ again unless you're taking limits of real (or complex) numbers. Particularly, if you take limits of sets, use ordinals instead.
The reason is that $\bigcup_{i=0}^\infty A_i$ is not properly defined in Set Theory: it is unclear what $\infty$ means, and there's no Axiom telling you that $\bigcup_{i=0}^\infty A_i$ is a set. The Axiom of Union tells you that if $\mathcal A$ is a set of sets, then $\bigcup \mathcal A$ is a set. If we index the elements of $\mathcal A$ with some index set $I$ as $\mathcal A=\{A_i\mid i\in I\}$, then we can use the alternative notation $\bigcup\mathcal A=\bigcup_{i\in I}A_i$. But, the index set needs to be a set for this union to be properly defined. By using $\infty$ you are not being clear what your index set is.
In case of $U(X)$, we define $S_0=X$, and for each $n\in\Bbb N$ we define $S_{n+1}=S_n\cup\mathcal P(S_n)$, and finally $U(X)=\bigcup_{n\in\Bbb N}S_n$.
$U(X)$ does not contain all von Neumann ordinals. It cannot contain all ordinals, since $U(X)$ is a set, and the von Neumann ordinals are a proper class. Not every von Neumann ordinal is hereditarily finite (and quite spectacularly so). For example, $\omega$ is not finite, $\omega_1$ is not even countable.
$V_{\omega+\omega}=U(V_\omega)$ and $V_\omega=U(\varnothing)$ are simply a consequence of writing out the definition of both sides and using that $V_\alpha\cup\mathcal P(V_\alpha)=V_{\alpha+1}$ for any ordinal (prove this, using transfinite induction).
So, to prove that $U(V_\omega)=V_{\omega+\omega}$, start by proving that $S_n=V_{\omega+n}$ for all $n\in\omega$. Then conclude that $$U(V_\omega)=\bigcup_{n\in\omega}S_n=\bigcup_{n\in\omega}V_{\omega+n}=V_{\omega+\omega}.$$
To prove that $U(\varnothing)=V_\omega$ is done similarly.
Another useful thing to prove is that if $X\subseteq Y$, then $U(X)\subseteq U(Y)$. Start by proving that $X\subseteq Y$ implies $\mathcal P(X)\subseteq \mathcal P(Y)$, and thus that $X\cup\mathcal P(X)\subseteq Y\cup \mathcal P(Y)$. You can then prove that $U(X)\subseteq U(Y)$ by induction.
From this it is immediate that $V_\omega\subseteq U(X)$ for any $X$, since $V_\omega=U(\varnothing)$.
This is false. I presume you confused a statement from my last answer, where I said that $\Bbb N$ "can be found" in $V_{\omega+1}$ and $\Bbb R$ in $V_{\omega+2}$.
What I mean precisely by this is the following. The usual way that Set Theory defines the set of natural numbers $\Bbb N$, is as the set of finite ordinal numbers $\omega$, or in other words, as the first limit ordinal. The relation on the natural numbers is given by the relation $\in$ defined on sets. In particular, the structure $(\Bbb N,<)$ as you know it is isomorphic to the structure $(\omega,\in)$. Moreover, operations such as addition and multiplication on natural numbers conveniently coincide with the addition and multiplication defined on the finite ordinals.
So, what is the least ordinal $\alpha$ such that $\omega\in V_\alpha$? First, note that $\omega$ is itself an ordinal. Second you can prove that $\alpha\subset V_{\alpha}$ and that $\alpha\in V_{\alpha+1}\setminus V_\alpha$ for every ordinal $\alpha$ (prove this by transfinite induction), so in particular
So the least $\alpha$ for which $V_\alpha$ contains the set $\omega$ of finite ordinals (i.e. natural numbers), is $\omega+1$. That is $\omega\notin V_\omega$, but $\omega\in V_{\omega+1}$. Note that $V_{\omega+1}$ contains a lot more sets than only $\omega$. For example, it also contains the set $\{(n,m)\mid n,m\in \omega\}$ of pairs of natural numbers (prove this, use that a pair $(n,m)=\{\{n\},\{n,m\}\}$ is a hereditarily finite set if $n$ and $m$ are hereditarily finite, and thus each pair is an element of $V_\omega$).
You can construct integers as the set $\omega\times \{0,1\}$, where we interpret $(n,0)$ as the integer $n$ and $(n,1)$ as the integer $-n-1$. You can show that $\omega\times\{0,1\}\in V_{\omega+1}$.
A rational number $\frac pq$ can be seen as a pair of integers. Again, you can show that the set of pairs of integers (as defined above) is also an element of $V_{\omega+1}$. That is, the rational numbers also are an element of $V_{\omega+1}$.
The real numbers $\Bbb R$ can be constructed in several different ways, but one way is as a pair of subsets of rational numbers (Dedekind cuts). Note that if $A$ is a set of rational numbers, then $A$ is a set of elements of $V_{\omega}$, and thus $A\in V_{\omega+1}$, and thus if $R$ is a family such that each $A\in R$ is a set of rational numbers, then $R\subset V_{\omega+1}$. Therefore $R\in\mathcal P(V_{\omega+1})=V_{\omega+2}$.
In this way, we have shown that $\Bbb R\in V_{\omega+2}$.
We can continue of course: functions $f:\Bbb R\to\Bbb R$ consist of pairs of real numbers, and thus are also elements of $V_{\omega+2}$. That means that the set of all real functions is an element of $V_{\omega+3}$, and thus function spaces are members of $V_{\omega+3}$.
Finally we arrive at this question. What is meant by "usual" is intentionally kept vague. Within foundational research areas, such as set theory or category theory, the set $V_{\omega+\omega}$ is not enough to work with. However, for the majority of mathematics, in areas like number theory, topology, geometry, probability, statistics, calculus, dynamics, etc. you will be working with things like the natural numbers, the real or complex numbers, function spaces, perhaps even function spaces of functions. It is very rare to encounter an object in these fields of research that cannot be modelled by some set contained in $V_{\omega+n}$ for $n<10$, and even more rare to encounter an object not contained in $V_{\omega+\omega}$. That is not to say that this doesn't happen, but for the majority of research areas, you very rarely need anything outside of $V_{\omega+\omega}$.