Regarding the title, Cantor normal form applies to all ordinals, not just ordinals up to $\varepsilon_0$. Every ordinal can be uniquely expressed as a finite sum $\omega^{\alpha_n}+\cdots+\omega^{\alpha_0}$ with $\alpha_n\geq\cdots\geq\alpha_0$. Below $\varepsilon_0$, the interesting thing is that you can get a nice terminating finite hereditary representation, with the exponents also represented this way. At $\varepsilon_0$, however, the hereditary representation starts not to work, since the representation $\varepsilon_0=\omega^{\omega^{\omega^{\cdot^{\cdot}}}}$ doesn't terminate.

If you intend to ask about the structure whose domain is the ordinals below $\varepsilon_0$, then you should write $(\varepsilon_0,\ldots)$, since $\text{Ord}^{<\varepsilon_0}$ would usually denote the proper class of all sequences of (arbitrary) ordinals, with length less than $\varepsilon_0$. Similarly, $\omega$ is the set of finite ordinals, whereas $\text{Ord}^{<\omega}$ is the class of all finite sequences of ordinals.

@ZuhairAl-Johar You have an idiosyncratic meaning for $H_\kappa$, since when $\kappa$ is singular, this usually means the collection of sets whose transitive closure is less than $\kappa$, but this is evidently not what you are doing. Why not use the usual terminology and notation? Also, with just extensionality and separation, but not foundation, etc., I'm not entirely sure what it means to be "a stage of the cumulative hierarchy", since that theory simply does not prove that there is a robust theory of such stages, and so I don't know exactly how you intend to express the reflection axiom.

@AsafKaragila Those refer to set difference and negation. So a theorem by Karagila\Hamkins would mean that you proved it entirely on your own, despite harmful sabotaging suggestions by me.

My point is that the definition of the game itself (for any $n$ or $m$) is unclear until you say whether I can pick 0 items on the first move.

You haven't clarified whether choosing 0 is allowed in general. And you say, "we choose everything," but do we also choose who moves first? In this case, isn't it obviously a win? Just choose a trivial case that is winning. I seem to be missing what the question is here.

You say, "we can always win", but of course the fundamental theorem of finite games is that for each initial pile height and fixed $m$, exactly one of the players has a winning strategy. But obviously it won't always be the same player, since after making a winning move, the other guy faces a losing pile height. You can inductively compute which are winning this way. But I guess you want to pick both $n$ and $m$?

Isn't 2 winning for player 1, since I pick 0, and then you have to pick 1 or 2, but if you pick 1 I pick 0 again and you lose, and if you pick 2 you lose directly. Or did you mean you can't pick 0 except in the case nothing else is possible?

This is very nice.

@NoahSchweber It seems that in addition to that account with computable binary sequences, Lurie's paper also has an account of the computable surreals that follow my proposal here, and he made the connection with $L_{\omega_1}^{CK}$ and so forth. Much of what has been observed on this question and its answers also already appears in that paper.

The field $\text{No}(\omega_1)$ is initial amongst all countably saturated real-closed fields. I believe a similar property will hold of $\text{No}(\kappa)$ for $\kappa$-saturated real-closed fields for many $\kappa$.

@ChristopherKing One cannot define them by any second-order property, since $\beta$ can see that $\alpha$ has all the second-order properties that it has. See my paper with Robin Solberg at arxiv.org/abs/2009.07164 for a discussion of related issues, defining $V_\kappa$ rather than $\kappa$.

The inequality you state as "best we could show" is the converse of the inequality in the question. Is that what is meant?

To clarify, you assume $G$ and $H$ are surreal numbers, and not merely games?

Great question, and I'm glad you answered. No need to delete. It would be great, however, if you could add more explanation, to make the answer somewhat self-contained. Links to articles are great and valuable, but one must recognize that most readers will not follow the links. Often one can just explain the idea quickly, and that is the best.