How to find the number of atomic microstates for a given electronic configuration?
You really have to calculate out things exactly, and particularly carefully if electrons are in the same orbital, e.g. $p^2$, when the Pauli principle has to be checked. Its a bit tedious but it can be very quick with a bit of practice especially if electrons are in different orbitals. 
You start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number s = 1/2 and magnetic quantum number $m_s$ = +-1/2. Orbitals have angular momentum of s=0, p=1, d=2, f=3 etc units.
The total spin angular momentum is the series of values $S=|s1+s2|...|s1-s2| $ where s1 and s2 =+1/2 are the quantum numbers of each electron. The total angular momentum is   $L=|l1+l2|...|l1-l2| $  and total angular momentum $J=|L+S|...|L-S| $ 
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by 1. Note that sometimes there is only one value and all values are positive. e.g. J=5/3, 3/2, 1/2 or J could be 0 or 3 etc.
The term symbol has the form $^{2S+1}L_J$.
The super-prefix is the spin multiplicity, for spin angular momentum $S$ this is  $2S+1$ or in general $2X+1$ for angular momentum X.
 In your question $s^1f^1$ the total spin using the series above is  S = 0 and 1. The spin multiplicity is thus 1 and 3, singlets and triplets terms only. The total orbital angular momentum L = 3 since the s orbital has zero angular momentum. Thus the terms so far are 
$^1F$ and $^3F$ The number of microstates (multiplicity) in each term is $(2S+1)(2L+1)$. This is 7 for the $^1F$ and 3.7=21 for the $^3F$ making 28 in total. These states are also degenerate in energy. You should calculate that $s^1d^1$ has 20 microstates.

The total angular momentum J series associated with S=1 is $J=|3+1|.... |3-1|$ or 4, 3 and 2 and with S=0 , J=3.  Thus only triplet states have J = 4 and 2 and both singlet and triplets J=3. The terms symbols for the levels are 
$^1F_3$, $^3F_4$, $^3F_3$,  $^3F_2$ 
 The total multiplicity from the J values is is 7+9+7+5 = 28 the same as from terms.

When the electrons are in the same orbital then the only secure way of calculating terms is to list all the individual quantum numbers in a table, with columns $m_{l1},m_{l2},m_{s1},m_{s2}$, S and L and remove those for which the Pauli principle removes a row. (See table in Steinfeld 'Molecules &amp; Radiation' chapter 2 and Engel &amp; Reid, 'Physical Chemistry' chapter 22). In particular you have to look out for when two rows have the same set of quantum numbers but in a different order; for example $m_{s1}$=0, $m_{s2}$=1 in one row and $m_{s1}$=1 and $m_{s2}$=0 in another but the $m_l$ values are the same, and vice versa. Doing this you should find, for example that $p^2$ has 15 terms, as $^1D$, $^3P$ and $^1S$ Hope this rather long answer helps you understand things.