If he said that he multiplied the expression by $2x$, he misspoke. He multiplied it by $\frac{2x}{2x}$. Note that $\frac{2x}{2x}=1$, so that it’s entirely permissible to multiply by it, while multiplying by $2x$ would change the value.
He took a small shortcut. I’ll do it the long way first, putting the numerator over a common denominator and simplifying the resulting three-story fraction:
$\begin{align*} \frac{\frac12-\frac1x}{x-2}&=\frac{\frac12\cdot\frac{x}x-\frac1x\cdot\frac22}{x-2}\\\\ &=\frac{\frac{x}{2x}-\frac2{2x}}{x-2}\\\\ &=\frac{\frac{x-2}{2x}}{x-2}\\\\ &=\frac{\frac{x-2}{2x}}{x-2}\cdot\frac{2x}{2x}\tag{1}\\\\ &=\frac{x-2}{2x(x-2)}\;. \end{align*}$
The tutor merely observed that when the numerator is put over a common denominator, that denominator will be $2x$, and avoided the first few steps of my calculation by essentially going directly to the step marked $(1)$:
$\begin{align*} \frac{\frac12-\frac1x}{x-2}&=\frac{\frac12-\frac1x}{x-2}\cdot\frac{2x}{2x}\\\\ &=\frac{\left(\frac{x}{2x}-\frac2{2x}\right)2x}{2x(x-2)}\\\\ &=\frac{x-2}{2x(x-2)}\;. \end{align*}$
There’s no swapping of the denominator into the numerator: it just happens that when the numerator is simplified, the result is a fraction whose numerator is the same as the original denominator.