1. Interpret the function carefully

   The given function is $f(x)=\sin^{-1}(2x)+\sin 2x+\cos^{-1}(2x)+\cos 2x,\qquad x\in\left[0,\frac{\pi}{8}\right].$

   Here, $\sin^{-1}(2x)$ and $\cos^{-1}(2x)$ mean inverse trigonometric functions of the argument $2x$.

2. Use the standard identity

   For $t\in[-1,1]$, $\sin^{-1} t+\cos^{-1} t=\frac{\pi}{2}.$

   Since $x\in\left[0,\frac{\pi}{8}\right]$, we have $0\le 2x\le \frac{\pi}{4}<1,$ so the identity is valid.

   Therefore, $\sin^{-1}(2x)+\cos^{-1}(2x)=\frac{\pi}{2}.$

   Hence $f(x)=\frac{\pi}{2}+\sin 2x+\cos 2x.$

3. Reduce the problem to finding extrema of $\sin 2x+\cos 2x$

   Let $g(x)=\sin 2x+\cos 2x.$

   Since $x\in\left[0,\frac{\pi}{8}\right]$, we have $2x\in\left[0,\frac{\pi}{4}\right].$

   Now, $g(x)=\sqrt{2}\sin\left(2x+\frac{\pi}{4}\right).$

   As $2x+\frac{\pi}{4}$ varies from $\frac{\pi}{4}$ to $\frac{\pi}{2}$, and $\sin\theta$ is increasing on this interval, $g(x)$ is increasing on the whole domain.

   So:

   • minimum occurs at $x=0$
   • maximum occurs at $x=\frac{\pi}{8}$

4. Compute minimum value $m$

   At $x=0$, $f(0)=\frac{\pi}{2}+\sin 0+\cos 0=\frac{\pi}{2}+0+1=\frac{\pi}{2}+1.$

   Thus, $m=\frac{\pi}{2}+1.$

5. Compute maximum value $M$

   At $x=\frac{\pi}{8}$, $f\left(\frac{\pi}{8}\right)=\frac{\pi}{2}+\sin\frac{\pi}{4}+\cos\frac{\pi}{4}$ $=\frac{\pi}{2}+\frac{1}{\sqrt2}+\frac{1}{\sqrt2}=\frac{\pi}{2}+\sqrt2.$

   Thus, $M=\frac{\pi}{2}+\sqrt2.$

6. Find $m+M$

   $m+M=\left(\frac{\pi}{2}+1\right)+\left(\frac{\pi}{2}+\sqrt2\right)=\pi+1+\sqrt2.$

7. Match with the options

   $m+M=1+\sqrt2+\pi,$ which is Option A.

Conclusion: The correct answer is A.