Fate dice can be represented by this polynomial:
$$\frac{x^{-1} + 1 + x^1}{3}$$
or
$$\frac{1 + x^1 + x^2}{3x}$$
When you roll 4 dice you get:
$$\frac{(1 + x^1 + x^2)^4}{81x^4}$$
or
$$\frac{(1 + x^1 + x^2)^4}{81x^4}$$
or
$$\frac{{(1 + x^1 + x^2)^2}^2}{81x^4}$$
$$\frac{(1 + 2 x^1 + 3 x^2 + 2x ^3 + x^4)^2}{81x^4}$$
$$\frac{1 + 4 x^1 + 10 x^2 + 16 x ^3 + 19 x^4 + 16 x^5 + 10 x^6 + 4 x^7 + x^8}{81x^4}$$
$$\frac{1x^{-4} + 4 x^{-3} + 10 x^{-2} + 16 x ^{-1} + 19 x^0 + 16 x^1 + 10 x^2 + 4 x^3 + x^4}{81}$$
where the \$x^n\$ coefficient is the chance you'll get the result \$n\$ after the roll.
Adding +2 had no variance as you know exactly what the result will be; you are adding after you see the result.
Rerolling compared to just a +2 when you are at -2:
It has a \$\frac{5}{81}\$ chance of being worse, a \$\frac{10}{81}\$ chance of not doing anything, and a \$\frac{31}{81}\$ chance of being worse than adding +2.
It also has a \$\frac{31}{81}\$ chance of being better than your +2.
In comparison, when at -3:
It has a \$\frac{1}{81}\$ chance of being worse, a \$\frac{4}{81}\$ chance of not doing anything, and a \$\frac{15}{81}\$ chance of being worse than adding +2.
It also has a \$\frac{50}{81}\$ chance of being better than your +2.
When at -4:
It has no chance of being worse, a \$\frac{1}{81}\$ chance of not doing anything, and a \$\frac{4}{81}\$ chance of being worse than adding +2.
It also has a \$\frac{66}{81}\$ chance of being better than your +2.
...
But really, the player should calculate the expected difficulty distribution and subtract it from their roll. Then work out if +2 is better than a reroll.
A cute way to model this is generate a distribution of the expected difficulty, then subtract it from your skill bonus.
So if your skill bonus is +3 and you expect the difficulty to be 50% 0 and 50% 1, then this is the polynomial \$\frac{x^1 + x^2}{2}\$. Multiply that by the above FATE roll polynomial to get the full distribution of your chance of success on a reroll.
For the +2 case, you just apply your known roll (say \$x^{-3}\$) to the difficulty distribution and see what your odds are of success (net of \$x^0\$ or higher).
...
You could also think about the scenarios. If the difficulty is D, what are the odds that +2 wins? A reroll wins? A reroll with a +2?
Then work out the relative odds of each possible difficulty.
...
Naturally, doing this live at the table is not that polite, unless you are playing by snail mail post. So you end up having to wing it.
You'll use a reroll if you don't think +2 is likely to succeed, but a reroll (and possibly a +2 after that) is better odds for the cost.