Computing a cardinal and an ordinal
Let me rephrase the question a bit, since I think as written it's a bit harder to parse than necessary:
As you've defined it, the map $t$ just sends a chain in $2^\mathbb{N}$ to its cardinality; there's no need to refer to $a$ at all. So $t_0$ is the supremum of the cardinalities of chains in $2^\mathbb{N}$. Similarly, $w_0$ is the supremum of the ordertypes of well-founded chains.
Let's think about $t_0$ first. Clearly $t_0$ is at most $\mathfrak{c}$ (= the cardinality of $\mathbb{R}$); in fact, this upper bound sharp - $t_0=\mathfrak{c}$ - and is in fact achieved, so is a maximum and not just a supremum.
To show this we just need to find a single chain of cardinality $\mathfrak{c}$. The idea here is to use Dedekind cuts. Fix a bijection $f: \mathbb{N}\rightarrow\mathbb{Q}$, and for a real $r$ let $S_r=\{f^{-1}(q): q\le r\}$. Then the collection $\{S_r: r\in\mathbb{R}\}$ is a chain of size continuum.
Alright, now let's turn to $w_0$. Suppose I have a well-ordered chain of ordertype $\theta$, $S=\{X_\eta: \eta<\theta\}$ with $X_\eta\subseteq X_\zeta$ iff $\eta\le\zeta$. For simplicity let's assume $\theta$ is a limit ordinal; this will be good enough. Then for each $\eta<\theta$, there is some $n\in\mathbb{N}$ in $X_{\eta+1}\setminus X_\eta$; let $n_\eta$ denote the least such natural number. Since there are only countably many natural numbers, this means that $\theta<\omega_1$.
This tells us $w_0\le\omega_1$. It also tells us that $\omega_1$ can't be achieved, so if we want to show $w_0=\omega_1$ we'll have to work a bit harder: fixing $\theta<\omega_1$ we'll need to build a well-ordered chain of ordertype $\theta$.
But this is straightforward: it's a standard exercise that every countable linear order embeds into $\mathbb{Q}$ in an order-preserving way. Let $f$ be as above, fix an order-preserving $h: \theta\rightarrow \mathbb{Q}$, and let $X_\eta=\{f^{-1}(q): q\le h(\eta)\}$. Then $S=\{X_\eta: \eta<\theta\}$ is a well-ordered chain of ordertype $\theta$.
First note that $t_0$ has an obvious upper bound, the cardinality of $\mathcal P(\Bbb N)$, which happens to be $2^{\aleph_0}$. But we also know that there is a chain in $\mathcal P(\Bbb N)$ which is isomorphic to the real numbers, so this upper bound is obtained, and therefore $t_0$ is exactly $2^{\aleph_0}$.
For $w_0$, note that every partially ordered set embeds into its power set (with inclusion), and two power sets are isomorphic (as ordered sets) if and only if the underlying sets have the same cardinality. This gives you a quick proof that every countable ordinal embeds into $\mathcal P(\Bbb N)$. So $w_0\geq\omega_1$.
But, if $W$ is a well-ordered chain, denoting by $A'$ the successor of $A$ inside $W$ gives us an injection $A\mapsto\min A'\setminus A$ from $W$ into the natural numbers (this might need minor adjustments for a maximal element of $W$, but that does not matter overall). So it means that no member of $\mathcal W$ is uncountable. Therefore $w_0=\omega_1$, the first uncountable ordinal.
Two remarks:
Your definitions are cumbersome. You can just talk about the cardinals or ordinals. No need to choose $a$ and $C$ and $\Omega$ and all that.
If you already know that every countable ordinal embeds into the real numbers, then by the argument from $t_0=2^{\aleph_0}$ you already have that every countable ordinal embeds into $\mathcal P(\Bbb N)$. There are other, similar, methods which may be easier depending on your prior knowledge.
Or perhaps a direct approach might be clearer to you. It's all the same, really.