would be the leading monomial of $f$. Write $m_i:={\mathrm{lm}}(b_i)$ for $i=0,1$, and let $t_0,t_1$ be such that ${\operatorname{lcm}}(m_0,m_1)=t_0m_0=t_1m_1$. Now $m={\mathrm{lm}}(f_{b_0})m_0={\mathrm{lm}}(f_{b_1})m_1$ is a multiple of both $m_0$ and $m_1$, and therefore ${\mathrm{lm}}(f_{b_0})$ is divisible by $t_0$; set $$A:=\frac{{\mathrm{lt}}(f_{b_0})}{t_0}.$$ Next consider $$S :=S(b_0,b_1) = t_0b_0-t_1b_1,$$ where we have used that $b_0$ and $b_1$ are monic. As $0$ is a remainder of $S$ modulo $B$ by assumption, we can write $S$ as a sum $\sum_{b \in B} s_{b}b$ with only finitely many non-zero terms that moreover satisfy ${\mathrm{lm}}(s_{b} b) \leq {\mathrm{lm}}(S) <
{\operatorname{lcm}}(m_0,m_1)$ for all $b$. Then we may rewrite $f$ as $$f=f-A(S - \sum_{b} s_{b} b) =
\sum_{b} (f_{b}+f'_{b}+f''_{b}) b$$ where $f'_{b} = A s_{b}$ and $$f''_{b} = \left\{
\begin{array}{ll}
-{\mathrm{lt}}(f_{b_0}) & \mbox{if $b= b_0$,} \\[2pt]
{\mathrm{lc}}(f_{b_0}){\mathrm{lm}}(f_{b_1}) &
\mbox{if $b = b_1$,} \\[2pt]
0 & \mbox{otherwise.}\\
\end{array}\right.$$ For all $b \in B$ we have $$\begin{aligned}
{\mathrm{lm}}((A s_{b})b)&={\mathrm{lm}}(A s_{b} b)
<\frac{{\mathrm{lm}}(f_{b_0})}{t_0} {\operatorname{lcm}}(m_0,m_1)\\
&= {\mathrm{lm}}(f_{b_0}) m_0=m,\end{aligned}$$ so for all $b$ we have ${\mathrm{lm}}(f'_{b} b) < m$. Moreover, ${\mathrm{lm}}((f_{b_0}+f''_{b_0})b_0)$ is strictly smaller than $m$. Finally, ${\mathrm{lm}}(f''_{b_1}b_1) = m$. We conclude that either $\max_{b}{\mathrm{lm}}((f_{b}+f'_{b}+f''_{b}) b)$ is strictly smaller than $m$, or else the number of $b$ for which it equals $m$ is smaller than the number 