To find the maximum value of the function $f\left(x\right)=\sin 2x+\cos 2x$ on the interval $[0,\frac{π}{2}]$, we can use trigonometric identities and properties of trigonometric functions.

First, we express the function in a more convenient form using the sum-to-product identity:
$$f\left(x\right)=\sin 2x+\cos 2x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin 2x + \frac{1}{\sqrt{2}}\cos 2x\right)$$
Recognizing that $\frac{1}{\sqrt{2}} = \cos\frac{π}{4}$ and $\sin\frac{π}{4} = \frac{1}{\sqrt{2}}$, we can rewrite the function as:
$$f\left(x\right) = \sqrt{2}\left(\sin\frac{π}{4}\sin 2x + \cos\frac{π}{4}\cos 2x\right)$$
Using the angle sum identity for sine, we get:
$$f\left(x\right) = \sqrt{2}\sin\left(2x+\frac{π}{4}\right)$$

Next, we determine the range of $2x+\frac{π}{4}$ given $x∈[0,\frac{π}{2}]$:
$$2x+\frac{π}{4}∈\left[\frac{π}{4},\frac{5π}{4}\right]$$
This range includes the angle $\frac{π}{2}$, at which the sine function reaches its maximum value of $1$.

Therefore, when $2x+\frac{π}{4}=\frac{π}{2}$, the function $f\left(x\right)$ reaches its maximum value, which is:
$$f\left(x\right) = \sqrt{2}\sin\left(\frac{π}{2}\right) = \sqrt{2}$$
Thus, the maximum value of the function $f\left(x\right)$ on the interval $[0,\frac{π}{2}]$ is $\boxed{\sqrt{2}}$.