Let's address points 2 and 3 first.
2: The empty game is the game where neither player has any legal move. The authors of that article are (implicitly, although they don't say it, it applies to all of their examples) only considering games under the normal (as opposed to misere) play convention, where the last player to move wins.
3: Let's make sure you have correct definitions of winning and losing positions. A winning position is a position where the side to move can win, and a losing position is a position where the side to move loses no matter what they do. To show a position is winning, I just need to exhibit a single winning move. It must lead to a position where the opponent loses no matter how they respond, i.e., the new position is a losing position for the opponent who is now on move.
Now 1 follows by an inductive argument from 2 and 3, assuming that (1) the game has finitely many positions; (2) it is impossible to repeat a position; (3) the game is impartial, i.e., the two players have the same legal moves. (These assumptions are not stated, but applies to all their examples.) If (3) did not hold, there would be four types of positions, not just two (winning and losing).
The inductive argument runs like this: By the additional assumptions above, the game can be represented as a finite tree where the starting position is at the root, and the leaves are all terminal positions where neither player can move. Argue by induction on the height of the nodes. At height 0 (the leaves), we have the empty position which we know is losing. At any higher node, the children are already classified as winning or losing (by induction). If all the children are winning, then the position is losing; otherwise, at least one child is losing, so the position is winning. Therefore, all nodes can be classified as winning or losing.
Here is a simple example of a "computable but non-computably-determined" game. Even better, each play lasts at most three moves. Below, "$\Phi_e(c)[s]\uparrow$" (respectively, $\Phi_e(c)[s]\downarrow$) denotes "the $e$th Turing machine on input $c$ does not halt in $\le s$ steps" (respectively, "does halt in $\le s$ steps:)
Player $1$ plays some natural number $n$.
Player $2$ responds with either a natural number $m$ of their own, or the word $\mathsf{never}$.
If player $2$ played a natural number, then the game stops; if player $2$ played the word $\mathsf{never}$, then player $1$ plays a natural number $k$.
Player $1$ wins iff $(i)$ player $2$ played a natural number $m$ but $\Phi_n(n)[m]\uparrow$, or $(ii)$ player $2$ played $\mathsf{never}$ but $\Phi_n(n)[k]\downarrow$. Player $2$ wins otherwise. Note that the rules and win conditions are computable. Also, there are no ties or infinite plays, so "non-losing strategy" is the same as "winning strategy" here (which simplifies things a bit).
Essentially, in this game player $2$ is trying to guess the halting problem. Obviously player $1$ has no winning strategy, computable or otherwise; however, any winning strategy for player $2$ has to be non-computable (and indeed must compute the halting problem).
Note that there is a kind of "infinite branching" here: specifically, the game tree is infinitely branching since there are infinitely many legal moves. This is necessary if we want to keep our game of bounded length: any computable game on a finite move set where all plays have length $\le n$ for some fixed $n$ must have a computable winning strategy (just via brute-force analysis).
Something a bit more interesting happens if we require a finite move set but don't require a uniform bound on the length of plays, and merely require that all plays be finite. While Konig's lemma implies that any such game does in fact have a uniform bound on the length of plays, that result is not computably true. On the other hand, it's also not as powerful as the halting problem (in a precise sense).
To make the picture somewhat more complete, there are four different ways we might make sense of the claim "Every finite-length game is determined," and these give rise to four different "levels of non-computability" as measured by reverse mathematics. This is all extremely technical, but I think it's worth putting these facts down for ease of reference even if they're getting beyond the context of this question:
"Every uniformly-finite-length game on a finite move set is determined" is at the lowest level $\mathsf{RCA_0}$ (= no noncomputability at all).
"Every non-uniformly-finite-length game on a finite move set is determined" is one level up, at $\mathsf{WKL_0}$.
"Every uniformly-finite-length game on an infinite move set is determined" is one level above that, at $\mathsf{ACA_0}$.
- Actually I'm lying - that's only true if we restrict attention to $\omega$-models. Properly speaking the right equivalent is the slightly-stronger system $\mathsf{ACA_0'}$ = $\mathsf{RCA_0}$ + "For each set $X$ and each number $n$, the set $X^{(n)}$ exists."
Finally, "every non-uniformly-finite-length game on an infinite move set is determined" - called clopen determinacy - is even higher, at the level of $\mathsf{ATR_0}$.
Infinitely-long games are also extensively studied, but more so in set theory than in reverse mathematics. That said, one of the earliest serious results in reverse mathematics was the surprising equivalence between clopen determinacy (= $\mathsf{ATR_0}$) and a particular infinite-length determinacy principle (basically, player $2$ wins if the game goes on forever). I've said a bit about this here.
Finally, this blog post of Hamkins may be of interest.
Best Answer
Essentially, no. For one thing, you can only take a product in a general way that works out at all if the game is cold (so that the game/summands act like numbers).
This is very contrived, but I suppose you could take a cold game that breaks up into sums, and use the definition of the exponential function in the Surreals to define the exponentiated game, which would break up as a product of the exponentials of the summands of the original. But if you're looking at some common game like Go, or even the contrived chess puzzles Noam Elkies applied CGT to, you're not going to have a shot at using "products" in any way.