Floating point precision from rearranging equation
First of all, the different numbers of digits are misleading. The two numbers store the same amount of precision; Python is just leaving off final zeroes.

However, there is a loss of precision here, and it comes down to the intermediate results.

Floating-point values aren't infinitely precise, and the larger their absolute value, the fewer places they can store after the decimal point. (Hence the name "floating point": one way to think of floating point numbers is that they're a fixed number of digits long, but you can move the decimal point back and forth to represent both really big and really small values.)

When you compute 8.8888888888**2, you're effectively using more of the available bits to store things before the decimal point. So there are fewer left over to store things after the decimal point. It gets less precise.

When you instead compute 8.8888888888 - 9.9999999999, you're not using those extra bits before the decimal point, so they can all be devoted to the part after the decimal point.

(Note: this is an intuitive explanation, not a technical one. The essence is correct, but the details are somewhat handwaved. If you want more technical details, look into how an IEEE float is stored.)