Why is the stoichiometric coefficient in the denominator in the rate expression?
From the stoichiometric coefficients $\nu_{i}$ in your reaction equation

$$\ce{2A + B → C}$$

you can see (as you already correctly stated) that for each molecule of $\ce{B}$ consumed, $\nu_{B} = -1$, two molecules of $\ce{A}$ must be consumed as well, $\nu_{A} = -2$. Similarly for each molecule of $\ce{B}$ consumed one molecule of $\ce{C}$ must be formed, $\nu_{C} = +1$ (and analoguously for $\ce{A}$ and $\ce{C}$).
This will also be reflected in the changes of the amounts of those species, $\Delta n_{i}$, during the reaction, e.g. if the amount of $\ce{B}$ changes by $2 \, \mathrm{mol}$ you can expect a change in the amount of $\ce{A}$ by $4 \, \mathrm{mol}$. So, the ratios of the $\{\Delta n_{i}\}$ and the $\{\nu_{i}\}$ will be equal, i.e.

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{B}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{B}}} \ ; &amp;&amp; \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{C}}} \ ; &amp;&amp; \frac{\Delta n_{\ce{B}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{B}}}{\nu_{\ce{C}}}
\end{align}

Rearranging these equations a bit leads you to:

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\nu_{\ce{A}}} = \frac{\Delta n_{\ce{B}}}{\nu_{\ce{B}}} = \frac{\Delta n_{\ce{C}}}{\nu_{\ce{C}}} \ .
\end{align}

Now, divide these equations by the reaction volume $V$ and use $\frac{\Delta n_{i}}{V} = c_{i} \equiv [i]$ to introduce concentration changes

\begin{align}
  \frac{\Delta [\ce{A}]}{\nu_{\ce{A}}} = \frac{\Delta [\ce{B}]}{\nu_{\ce{B}}} = \frac{\Delta [\ce{C}]}{\nu_{\ce{C}}} \ .
\end{align}

Finally, if you divide now by the reaction time $\Delta t$ and use the values for the stoichiometric coefficients, you get the equations from your question:

\begin{align}
  \frac{1}{\nu_{\ce{A}}} \frac{\Delta [\ce{A}]}{\Delta t} &amp;= \frac{1}{\nu_{\ce{B}}} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{\nu_{\ce{C}}} \frac{\Delta [\ce{C}]}{\Delta t} \\
  \frac{1}{-2} \frac{\Delta [\ce{A}]}{\Delta t} &amp;= \frac{1}{-1} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{1} \frac{\Delta [\ce{C}]}{\Delta t} \\
-\frac{1}{2} \frac{\Delta [\ce{A}]}{\Delta t} &amp;= -1 \frac{\Delta [\ce{B}]}{\Delta t} = 1 \frac{\Delta [\ce{C}]}{\Delta t} \ .
\end{align}