The strategy I employ is simply to make the move that leaves the most available moves, and to disregard score entirely. As a natural consequence of playing more moves, the score of the board will increase simply because the only way to continue playing is to make combinations, and combinations generate higher scores.
At the beginning of the game, the most important thing to consider is the placement of $1$s and $2$s. They are unique in that nothing can be combined adjacent to them to make a valid combination, they will only combine with their complement, which can only be achieved by board translation (verses, say a
$12$, which can be adjacent to a $6$, which combines with another $6$ and then the $12$ can subsequently combine with that $12$. There's no way to make a $1$ or a $2$, it must simply be moved around the board).
Later, with higher scoring tiles on the board, the "bonus" tile (which shows up as a white tile with a $+$ in it at the top) becomes increasingly important, and the best strategy I've found is to attempt to place the bonus tile as near a mixed group of larger tiles as possible. The bonus tile will always be at least $6$, but never the same score as your highest scoring tile in play.
There is also the nature of the tile selection. It's been reverse engineered that the random generator uses a "bag" where $12$ tiles are shuffled. The original board layout uses this method, and $9$ tiles are placed into the board with $3$ remaining in the "bag". Once the bag is exhausted, the tiles are shuffled again. There are always $4$ of each: $1$, $2$, and $3$. Once you reach a high tile of $48$ a "bonus" tile is inserted with a potential value of greater than $3$. This changes the size of the "bag" to $13$ instead of $12$. So, keeping track of where you are in the "bag" and how many of each color you've seen can give you an advantage when looking at future moves.
Curiously, the possibility space for scoring is actually quite sparse. All scores will necessarily be multiples of $3$, but it turns out that only about $\frac38$ of the multiples of $3$ between $0$ and the max score are actually valid. There are a lot that are simply impossible to get, like a $19$ in cribbage.
The lowest one that isn't trivially small is still $39,363$, though, which seems well out of the range of the average player. The next lowest I found is $52,485$. There are lots of gaps at the high end, due to the fact that highest scoring tile is worth over $500$k by itself.
Best Answer
Based on your comment, you are really counting the number of single moves possible in white's first two. Each pawn has three possiblities, one space, two spaces, and (if it was advanced two on the first move) from rank 4 to rank 5, for a total 24. Each knight has two for a first move and eight for a second move (including back where it started-is 1 Na3 2Nb1 different from 1Nc3 2Nb1? Does that count as four moves?), total 20. Each rook has two for a second move. Each bishop has seven. The queen has nine and the king three. Overall total $24+20+2\cdot 2 + 2 \cdot 7 + 9+3=74$ possible moves in white's first two. It depends also whether you think 2Bh6 is different from 2BxNh6 and 2Bxh6. I have counted it only once.