1. Use the relation between inverse sine and inverse cosine

Let $A=\sin^{-1}x,\qquad B=\cos^{-1}x.$ Then $A+B=\frac{\pi}{2}.$

Given $\frac{\sin^{-1}x}{\alpha}=\frac{\cos^{-1}x}{\beta},$ so $\frac{A}{\alpha}=\frac{B}{\beta} \implies A\beta=B\alpha.$ Hence $\frac{\alpha}{\beta}=\frac{A}{B}.$ Therefore, for some constant $k$, $\alpha=kA,\qquad \beta=kB.$ So, $\alpha+\beta=k(A+B)=k\cdot \frac{\pi}{2}.$

2. Simplify the angle

We need $\sin\left(\frac{2\pi\alpha}{\alpha+\beta}\right).$ Substitute $\alpha=kA$ and $\alpha+\beta=k\frac{\pi}{2}$: $\frac{2\pi\alpha}{\alpha+\beta}=\frac{2\pi(kA)}{k\pi/2}=4A.$ Thus, $\sin\left(\frac{2\pi\alpha}{\alpha+\beta}\right)=\sin(4A).$ Since $A=\sin^{-1}x$, we have $\sin A=x,\qquad \cos A=\sqrt{1-x^2}.$ (Here $0<x<\frac{1}{\sqrt2}$ ensures $A$ is acute, so cosine is positive.)

3. Compute $\sin 4A$

Using $\sin 4A=2\sin 2A\cos 2A,$ we first find: $\sin 2A=2\sin A\cos A=2x\sqrt{1-x^2},$ $\cos 2A=1-2\sin^2A=1-2x^2.$ Therefore, $\sin 4A=2\cdot \left(2x\sqrt{1-x^2}\right)(1-2x^2).$ So, $\sin 4A=4x\sqrt{1-x^2}(1-2x^2).$

4. Match with the options

This matches Option B: $4x\sqrt{1-x^2}(1-2x^2).$

5. Final answer

$\boxed{4x\sqrt{1-x^2}(1-2x^2)}$ So the correct option is B.