Coefficient of relationship and path of coefficient
The expression $f = 2h-1,$ coefficient of inbreeding (COI) is a measure of homozygosity. Since in a randomly breeding herd (of cattle) we expect 50 per cent heterozygous and 50 per cent homozygous, the minimum for $2h-1 $ is 0. If the cattle's genome is purely homozygous (aa, AA) we have $2h-1 = 1.$ (h is total homozygosity--see Wright's paper, linked below.)  

As we can see from this site, a basic calculation of relationship coefficient may not deal with f at all. For two descendants B, C, the quantities $f_B,f_C$ are an attempt to factor in existing relative homozygosity that the coefficient of relation (COR) would otherwise not capture. Likewise $f_A$ measures preexisting homozygosity in the common ancestor of B and C. If animal A is the product of siblings then the homozygosity could already be quite high and this would maybe have measurable effects on the vitality of a cross between B and C.

Now $f_A$ and $f_B$ vary between 0 and 1 so $s = (\frac{1+f_A}{1+f_B})^{1/2}$ varies between $(1/2)^{1/2}$ and $(2)^{1/2}.$ Note that the important thing here is relative homozygosity of an animal and its ancestor:

a. If the ancestor A and descendent B have the same known preexisting burden of heterozygosity we can disregard s (and f) entirely. Their COR will depend only on path distance. 

b. The homozygosity of the ancestor A  tends to increase COR while that of B tends to decrease COR.  

The expression $r_{BC }= \sum p_{AB }\cdot p_{AC}$ is somewhat confusing. On page 334 of one of Wright's earlier papers we can get a clear idea of the use of the coefficient. 

He defines $p_{BA} = (1/2)^n (\frac{1+f_A}{1+f_B})^{1/2}.$ If the path from C to A is $p_{CA} = (1/2)^m (\frac{1+f_A}{1+f_C})^{1/2}$ then the total path BAC would be  

$$ p_{AB,AC} = (1/2)^{n+m}(\frac{1+f_A}{1+f_B})^{1/2}(\frac{1+f_A}{1+f_C})^{1/2} $$

$$= (1/2)^{n+m}\frac{1+f_A}{[(1+f_B)(1+f_C)]^{1/2}}  $$

So the expression for the sum of such paths would be 

$$r_{BC} = \sum_i p_{AB,AC} = \sum_i (1/2)^{k_i}\frac{1+f_{A_i}}{[(1+f_B)(1+f_C)]^{1/2}}   $$ 

in which $A_i$ is the ith ancestor through whom B and C are related. The total path length for a given term is $(n_i+m_i)$ which I abbreviated $k_i.$ 

So a less confusing expression for $r_{BC}$ might be: 

$$r_{BC }= \sum_i p_{A_iB}p_{A_iC}. $$

Note that the sum is over all common ancestors back to some arbitrary generation. 

A good example of this is Figure 11 at link 1 above. 
It's really not ``dumbed down'' at all, it just leaves out the factor s, which for a carefully maintained herd may be close to 1 anyway.