Recall that a set $X \subseteq \omega^\omega$ is determined if the Gale-Stewart game for $X$ is determined. It's well-known that the axiom of choice implies the existence of an undetermined set. The standard proof involves ordering all possible strategies of player I and II, then constructing $X$ by diagonalising all such strategies.
However, I feel that I don't have a good understanding of what this $X$ really look like. I would like to know if there is a more "explicit" way of constructing an undetermined set $X$. A good analogue would be how axiom of choice is used "more explicitly" to construct a non-Ramsey subset of $X \subseteq [\omega]^\omega$: Define an equivalence relation $\sim $ on $[\omega]^\omega$ by $A \sim B$ iff $|A \, \triangle \, B|$ is finite, and define $X = \{A \in [\omega]^\omega : |A \, \triangle \, A^*| \text{ is even} \}$, where $A^*$ is a chosen representative in the equivalence class $[A]$.
To put my question more specifically: Can we construct an undetermined set $X$ using an "explicit" definition (like the definition of the non-Ramsey set above), where the definition does not involve diagonalising all strategies of both players?
I think this is an interesting question, but one which is based on a subjective distinction which is difficult to make precise. That being said, I hope the following remarks are helpful:
First, note that each strategy $\Sigma$ (for either player) determines a perfect set $P_\Sigma\subseteq\omega^\omega$, namely the set of reals which can arise as plays consistent with $\Sigma$. A set being undetermined just means that it meets but does not contain every $P_\Sigma$. In particular, Bernstein sets (in the $\omega^\omega$-sense rather than the $\mathbb{R}$-sense) are undetermined; conversely, while undetermined sets need not be Bernstein, the "gamification" of the perfect set property (see Kechris) means that if you can produce an undetermined set then you can produce a Bernstein set. So this gives us a way to remove the game language from the subject, which may make it more intutitive.
Next, let me highlight what I think is the salient distinction between Vitali-style constructions and transfinite recursions. For $S$ a transitive model of enough set theory (say KP + Powerset), say that an $S$-transversal is a set $X\in S$ which $S$ thinks is a transversal for the Vitali equivalence relation. The point is that transversals amalgamate well: if $M, N_1, N_2, K$ are transitive models of enough set theory with $M\subseteq N_1\cap N_2 \subseteq N_1\cup N_2\subseteq K$ then
for every $M$-transversal $S$ there are $N_1, N_2$-transversals $T_1,T_2$ respectively with $S=T_1\cap M=T_2\cap M$;
for every $N_1$-transversal $T$, the set $T\cap M$ is an $M$-transversal (and similarly with $N_2$); and
if $T_1, T_2$ are $N_1,N_2$-transversals respectively with $T_1\cap M=T_2\cap M$ then there is a $K$-transversal $U$ with $T_1=U\cap N_1$ and $T_2=U\cap N_2$.
This means that even though choices are necessary in the construction of a transversal, they're somehow "model-stable" - the process of picking a representative for a given class doesn't look too much at the ambient model. (A relevant term here is pinned equivalence relations.) By contrast, the ordering-based nature of transfinite recursions makes the analogous result at least non-obvious.
That said - and this is the part which may actually be an answer to your question - there is such a way to build an undetermined set! This is what Mitchell Spector suggests in his comment: we have a canonical way to transform a nonmeasurable set into an undetermined game (see e.g. these notes of Rosendal or Martin's A simple proof that determinacy implies Lebesgue measurability), and we have per the above a "model-stable" way to produce a nonmeasurable set.