User [hexomino](https://puzzling.stackexchange.com/users/18422/hexomino) already [figured out the puzzle](https://puzzling.stackexchange.com/a/87581/36023), and managed to actually find the very complicated path that was exactly how I came up with the game. To recap:

>! you can interpret the game position as a binary number (marbles are ones, empty bowls are zeroes), and then every possible move becomes a (binary) subtraction of a power of two. This reduces the puzzle to [another game we have already solved](https://puzzling.stackexchange.com/q/86574/36023), so we are done.

The game itself is a lot easier to play than that, though, so I'm posting this self-answer to show how.

First, it's very useful to note that in a given board position, **every move is uniquely defined by the highest numbered bowl it affects**. So instead of saying `Take a marble from 2 and add a marble to bowls 3-7 each`, we can just say `play at 7`, and there's going to be exactly one way to do that. Not only is this a nice shorthand way to mark down moves, turns out it's quite crucial in solving the entire puzzle. **Below, all moves are given using this notation.** 

> <sup>(Now that we've established the notation, you can take a peek at the final spoiler block if you want to see the surprisingly simple conclusion at this point.)</sup>

The other, more significant part of the solution is

>! The balance of marbles in odd and even numbered bowls. Let's call that balance $\Delta$ (Delta). To calculate $\Delta$, simply count the marbles in odd-numbered bowls and subtract the number of marbles in even-numbered bowls.

To see why $\Delta$ so important, let's take a look at how the possible moves affect it:

>! A play on an **odd-numbered** bowl with a **marble** in it removes one odd bowl from the count, so **$\Delta$ decreases by one**.  
>!  
>! A play on an **odd-numbered** **empty** bowl, with a marble in the preceding bowl adds one odd bowl, and removes an even bowl, so **$\Delta$ increases by two**.  
>!  
>! A play on an **even-numbered** bowl with a **marble** in it removes one even bowl from the count, so **$\Delta$ increases by one**.  
>!  
>! A play on an **even-numbered** **empty** bowl, with a marble in the preceding bowl adds one even bowl, and removes an odd bowl, so **$\Delta$ decreases by two**.  
>!   
>! A play on **any empty bowl** with **more empty bowls** before it will be equivalent to one of the cases above: if a move affects two adjacent empty bowls, they will cancel each other out, and the remaining highest numbered bowl will have the same parity as the originally played bowl.

Pondering this for a while, we can find these helpful facts:

>! If we **divide $\Delta$ by 3**, the **remainder** will always  
>!   * increase by 1, if we play on an odd-numbered bowl.  
>!   * decrease by 1, if we play on an even-numbered bowl.

Since we know that the final, winning move leaves a $\Delta$ of exactly zero, and we know that any move either adds one to the remainder $\Delta$ (mod 3), or removes one from it, we can deduce that

>! * If $\Delta$ is divisible by three, there are no moves that leave $\Delta$ divisible by three, so  
>! * There are no winning moves from a position where $\Delta$ is divisible by three, and  
>! * If $\Delta$ is not divisible by three, there is always a move that leaves a $\Delta$ divisible by three for the opponent.

This gives the winning strategy:

>! Always play a move that makes $\Delta$ divisible by three.

Using this, we can judge the starting position:

>! There's one marble in an odd bowl, and one in an even bowl, so $\Delta$ is zero, and it's impossible to play a winning move, so the position is losing

so the (fastest) way to win is to

>! **let the opponent start**. Then, whenever the opponent plays on an even-numbered bowl, respond with an odd-numbered bowl (the smallest possible one), and vice versa.