Because the method of partial fractions says that the form of the denominator dictates the form of the numerators.
In your case, the denominator has factors of $s^2$ and $s+1$. The $s^2$ factor is referred to as a repeated linear factor (think $s^2=s\cdot s$ and $s$ is of course a linear factor since it is of the form $Cs+D$). On the other hand, $s+1$ is a distinct (nonrepeated) linear factor.
Looking in your text, Wikipedia, etc. will show that the method of partial fractions says that when you have a distinct linear factor, the numerator is just a constant. When you have a repeated linear factor that is repeated $k$ times, then you need to "count up" on powers of that factor until you reach its $k$th power. The numerator of each of those terms needs to be a distinct (that's important) constant.
That's why in your case, $f(s) = \underbrace{\frac{A}{s^2}+\frac{B}{s}}_{\text{repeated linear factor: ``counting up" on powers of }s\text{ (read from right to left)}}+\underbrace{\frac{C}{s+1}}_\text{distinct linear factor}$.
Finally, the method of partial fractions also says that if you have an (distinct) irreducible quadratic factor (i.e., it can't be factored over the reals), then you get a general linear function in the numerator rather than just a constant. For example, if you had ${1\over s^2(s+1)(s^2+1)},$ then the first two parts work like before, but the new factor in the denominator is an irreducible quadratic, so ${1\over s^2(s+1)(s^2+1)}={A\over s}+{B\over s^2}+{C\over s+1}+{Ds+E\over s^2+1}.$
Most generally, you can keep playing this game by allowing repeated irreducible quadratic factors, e.g. $(s^2+1)^3$, etc. But I think you have enough to go on...