To solve the given problem, we start by breaking down the trigonometric functions into their exact values:

1. $\tan 60^{\circ} = \sqrt{3}$
2. $\sin 45^{\circ} = \dfrac{\sqrt{2}}{2}$
3. $\cos 30^{\circ} = \dfrac{\sqrt{3}}{2}$

Now, substituting these values into the original expression:

$\tan 60^{\circ}+2\sin 45^{\circ}-2\cos 30^{\circ} = \sqrt{3}+2\times \dfrac{\sqrt{2}}{2}-2\times \dfrac{\sqrt{3}}{2}$

Simplifying the expression step by step:

$= \sqrt{3} + \sqrt{2} - \sqrt{3}$

Notice that $\sqrt{3}$ and $-\sqrt{3}$ cancel each other out, leaving us with:

$= \sqrt{2}$

Therefore, the final answer is $\boxed{\sqrt{2}}$.