where the ones on the left going downward are from $9\times1$ to $9\times5$, and the ones on the right going upward are from $9\times6$ to $9\times10$. (This gives an insight into why.) When I was very young, I had difficulties remembering the multiplication tables; after being told of this property, I figured I could "fold over" the 9's table in half and get away with just remembering the first five, and that part of the multiplication tables subsequently became less difficult for me.
Take heed also of lhf's comments. Any multiple of 9 should have its digits sum to 9 or a multiple of 9. Even if you extend the table to, say, $9\times 13=117$, the digits of the answer should still sum to 9 or a multiple: $1+1+7=9$.
When there is equal precedence, the standard thing to do is to work from left to right. Hence
$$80/10\times 2 = (80/10)\times2=16. $$
It should be noted that this is not completely universal, i.e. some programming languages don't do it this way, as far as I know. Anyway, it doesn't really matter that much because no one actually writes $80/10\times2$, instead they write $(80/10)\times2$ to remove any ambiguity.
Best Answer
The sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$
Then the next row would be $(2+4+6+8+10+12+14+16+18) = 2\times45 = 90$
So the top two rows sum to $(1+2)\times 45 = 135$
Then it becomes obvious that the full sum of the products is the product of the sums, ie. $45\times45 = 2025$