We are given: $f(2)=4, \qquad f'(2)=4$ (The statement $f'(x)=4$ is understood here as the derivative at the required point $x=2$ for evaluating the limit.)

Evaluate the limit: $L=\lim_{x\to 2}\frac{x f(2)-2f(x)}{x-2}$

Substitute $f(2)=4$: $L=\lim_{x\to 2}\frac{4x-2f(x)}{x-2}$

Rewrite the numerator in a useful form: $4x-2f(x)=2\big(2x-f(x)\big)$ So, $L=2\lim_{x\to 2}\frac{2x-f(x)}{x-2}$

Now split: $2x-f(x)=\big(2x-4\big)-\big(f(x)-f(2)\big)$ since $f(2)=4$.

Therefore, $L=2\lim_{x\to 2}\frac{(2x-4)-(f(x)-f(2))}{x-2}$

Separate the limit: $L=2\left[\lim_{x\to 2}\frac{2x-4}{x-2}-\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}\right]$

Compute each part: $\lim_{x\to 2}\frac{2x-4}{x-2}=\lim_{x\to 2}\frac{2(x-2)}{x-2}=2$ and by definition of derivative, $\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=f'(2)=4$

Hence, $L=2(2-4)=2(-2)=-4$

So the correct option is: $\boxed{\text{C: }-4}$