Note: I'm using the general definition of BN-pair, which is weaker than the condition (B) given above.
It's a triangle inequality. One way to think about BN-pair is that they give a sort of combinatorial distance function on $G/B$. Given two cosets $g_1B$ and $g_2B$, you look at the product $Bg_1^{-1}g_2 B\in B\backslash G/B\cong N/(N\cap B):= W$ and think of that as the "distance" between them. To get a more numberish distance, you can let the length of an element of $W$ be the length of the shortest product of $x_i$'s which gives it.
If I take two cosets $BgB$ and $BhB$, and expand those as $Bx_{i_1}\cdots x_{i_n}B$ and $Bx_{j_1}\cdots x_{j_m}B$, then $$BgB\cdot BhB\subset Bx_{i_1}\cdots x_{i_n}B\cdot Bx_{j_1}\cdots x_{j_m}B\subset Bx_{i_1}B\cdots Bx_{i_n}Bx_{j_1}B\cdots Bx_{j_m}B.$$ Applying (B) inductively, we see that the last term is in the union of certain double cosets which have length shorter than the sum of that of $BgB$ and $BhB$.
To apply this to the "distance function," note that $Bg_1^{-1}g_3B\subset Bg_1^{-1}g_2B\cdot Bg_2^{-1}g_3B$, so the length of the distance between $g_1$ and $g_3$ is less than the sum of the lengths for $g_1$ to $g_2$ and $g_2$ to $g_3$: the triangle inequality.
Of course, that just shows that (B) implies the triangle inequality, but it's easy to see that it's also a special case.