After finding the possible values of $u$, solve $f(x)=u$ separately for each value.

Each equation for $x$ will have roots of the form $2+r$ and $2-r$. Multiply each pair using the difference of squares before multiplying the two pair-products.

Algebra is faster for an exact result, but Desmos can verify it. Enter $y=((x-2)^2-6)^2-4$, which is equivalent to $f(f(x))$, and enter $y=4$.

Click the four intersection points to read their x-coordinates. They are approximately $-0.971$, $0.219$, $3.781$, and $4.971$.

Multiply the four displayed x-coordinates in a new expression line. The result is approximately $-4$, confirming the exact algebraic result.

Let $u=f(x)$. Then $f(u)=4$, so $(u-2)^2-4=4$. Therefore, $(u-2)^2=8$, giving $u=2+2\sqrt{2}$ or $u=2-2\sqrt{2}$.

For each possible value of $u$, solve $f(x)=u$. If $u=2+2\sqrt{2}$, then $(x-2)^2=6+2\sqrt{2}$, so the two solutions are $2\pm\sqrt{6+2\sqrt{2}}$. If $u=2-2\sqrt{2}$, the two solutions are $2\pm\sqrt{6-2\sqrt{2}}$.

The product of the first pair is $4-(6+2\sqrt{2})=-2-2\sqrt{2}$. The product of the second pair is $4-(6-2\sqrt{2})=-2+2\sqrt{2}$. Thus, the product of all four solutions is $(-2-2\sqrt{2})(-2+2\sqrt{2})=4-8=-4$.