We use another, simpler, distributive law that holds in all complete Boolean algebras: $x \cdot \sum_i a_i = \sum_i (x \cdot a_i)$ which is exercise 17.15(a) in my copy of Jech.
This first law can be shown as follows: For any fixed $i$ we have that $a_i \le \sum_i a_i$, hence $x \cdot a_i \le x \cdot \sum_i a_i$, which makes the right hand side an upper bound for all $x \cdot a_i$, so $\sum_i (x \cdot a_i) \le x \cdot \sum_i a_i$
On the other hand, for all $a$ we have that a \le x' + a = x' + (a \cdot x), and applying this for all $a_i$ we get \sum_i a_i \le \sum_i (x' + (a_i \cdot x)) \le x' + \sum_i (a_i \cdot x), and now we multiply both sides by $x$: x \cdot (\sum_i a_i) \le x \cdot (x' + \sum_i (a_i \cdot x)) = (by finite distributivity!) = x\cdot x' + (x \cdot \sum_i (a_i \cdot x)) = x \cdot \sum_i (a_i \cdot x) \le \sum_i (a_i \cdot x), which put together gets us the required other inequality.
Now, to the case at hand: for every $i$ we have $u_{0,i} \cdot B_J = u_{0,i} \cdot \sum_j u_{1,j} = \sum_j (u_{0,i} \cdot u_{1,j}) \le v$ by applying the above law to the $B_J$ as the infinite sum, and noting that all products in that sum are from $B_{I \times J}$ and thus all bounded by the same $v$ (and so their sum is as well). But as this holds for all $i$: $\sum_i (u_{0,i} \cdot B_J) \le v$ and again by applying the above law again, pulling out the $B_J$: $\sum_i ( u_{0,i} \cdot B_J ) = \sum_i u_{0,i} \cdot B_J = B_I \cdot B_J,$ finally showing that $B_I \cdot B_J \le v$, as required.