1. Let $A=\cos^{-1}x,\qquad B=\cos^{-1}\left(\frac y2\right).$ Then the given condition is $A-B=\alpha.$ Also, $x=\cos A,\qquad \frac y2=\cos B \Rightarrow y=2\cos B.$

2. Substitute these into the expression: $4x^2-4xy\cos\alpha+y^2.$ Using $x=\cos A$ and $y=2\cos B$, \begin{align*} 4x^2-4xy\cos\alpha+y^2 &=4\cos^2A-4(\cos A)(2\cos B)\cos\alpha+(2\cos B)^2 \ &=4\cos^2A-8\cos A\cos B\cos\alpha+4\cos^2B \ &=4\left(\cos^2A-2\cos A\cos B\cos\alpha+\cos^2B\right). \end{align*}

3. Since $\alpha=A-B$, we use $\cos\alpha=\cos(A-B)=\cos A\cos B+\sin A\sin B.$ So inside the bracket, \begin{align*} \cos^2A-2\cos A\cos B\cos(A-B)+\cos^2B. \end{align*} But a cleaner way is to use the identity $\cos^2A+\cos^2B-2\cos A\cos B\cos(A-B)=\sin^2(A-B).$ Hence, $\cos^2A-2\cos A\cos B\cos\alpha+\cos^2B=\sin^2\alpha.$ Therefore, $4x^2-4xy\cos\alpha+y^2=4\sin^2\alpha.$

4. So the required value is $\boxed{4\sin^2\alpha}.$

5. Checking options:

• A: $2\sin2\alpha$ ❌
• B: $4$ ❌
• C: $4\sin^2\alpha$ ✅
• D: $-4\sin^2\alpha$ ❌

Thus the correct option is C.