To solve the given problem, we proceed as follows:

First, we express $2\cos 10^{\circ}$ in terms of sine using the co-function identity, which states that $\sin(90^{\circ}-\theta) = \cos\theta$ and vice versa. Thus, we have:
$$2\cos 10^{\circ} = 2\sin(90^{\circ}-10^{\circ}) = 2\sin 80^{\circ}.$$

Next, we use the sine addition formula, $\sin(a + b) = \sin a \cos b + \cos a \sin b$, to expand $\sin 80^{\circ}$ as $\sin(60^{\circ}+20^{\circ})$:
$$2\sin 80^{\circ} = 2\sin(60^{\circ}+20^{\circ}) = 2\left(\sin 60^{\circ}\cos 20^{\circ} + \cos 60^{\circ}\sin 20^{\circ}\right).$$

Substituting the exact values for $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$, we get:
$$2\left(\sin 60^{\circ}\cos 20^{\circ} + \cos 60^{\circ}\sin 20^{\circ}\right) = 2\left(\frac{\sqrt{3}}{2}\cos 20^{\circ} + \frac{1}{2}\sin 20^{\circ}\right) = \sqrt{3}\cos 20^{\circ} + \sin 20^{\circ}.$$

Therefore, substituting this result into the original expression, we have:
$$\frac{2\cos10°-\sin20°}{\cos20°} = \frac{\sqrt{3}\cos20°+\sin20°-\sin20°}{\cos20°} = \frac{\sqrt{3}\cos20°}{\cos20°} = \sqrt{3}.$$

Thus, the value of the given expression is $\boxed{\sqrt{3}}$.