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Substituting $y = \frac{x^2 - f(x)}{2}$ into the original equation, we get
$$f\left(\frac{x^2 + f(x)}{2}\right) = f\left(\frac{x^2 + f(x)}{2}\right) + 4\left(\frac{x^2 - f(x)}{2} - 2\right)(f(x) + 2).$$
Thus we obtain: $(x^2 - f(x) - 4)(f(x) + 2) = 0$; therefore, for every real number $x$, it always holds that
$$f(x) = x^2 - 4 \text{ or } f(x) = -2.$$

Furthermore, regardless of which of the above two cases holds, $f(\sqrt{2}) = -2$ always holds.

(1) We check that the function $f(x) = x^2 - 4$ satisfies the condition:
Left side: $f(f(x) + y) = f(x^2 - 4 + y) = (x^2 - 4 + y)^2 - 4,$
Right side: $f(x^2 - y) + 4(y - 2)(f(x) + 2) = (x^2 - y)^2 - 4 + 4(y - 2)(x^2 - 2)$.
By direct simplification, we obtain: Left side $-$ Right side $= 0$, therefore, $f(x) = x^2 - 4$ is one of the functions sought.

(2) If there exists some real number $a$ such that $f(a) \neq a^2 - 4$, then $f(a) = -2$, and $a^2 - 4 \neq -2$, that is, $a^2 - 2 \neq 0$.
We now prove that in this case, $f$ must be a constant function (for every real number $x$, $f(x) \neq -2$).
Substituting $x = a$ into the original equation, and using $f(a) = -2$, we obtain:
$$f(-2 + y) = f(a^2 - y).$$
On the other hand, substituting $x = \sqrt{2}$ into the original equation, and using $f(\sqrt{2}) = -2$, we obtain:
$$f(-2 + y) = f(2 - y).$$

Combining the above two equations, we get $f(a^2 - y) = f(2 - y)$.
Now substituting $z = 2 - y$ into this equation, we get $f(z) = f(z + a^2 - 2)$. Therefore, the function $f$ is a periodic function with period $p = a^2 - 2$. Hence, substituting $y = 0$ and $y = p$ respectively into the original equation, we know:
for every real number $x$, the following two equations both hold:
$$\begin{aligned} f(f(x)) &= f(x^2) - 8(f(x) + 2) \\ f(f(x) + p) &= f(x^2 - p) + 4(p - 2)(f(x) + 2). \end{aligned}$$
Since $p$ is a period of the function $f$, we have $f(f(x)+p) = f(f(x))$ and $f(x^2-p) = f(x^2)$;
from this we obtain: $(p-2)(f(x)+2) = -2(f(x)+2)$, that is, $p(f(x)+2) = 0$.
But $p = a^2 - 2 \neq 0$, so $f(x) = -2$ holds for every real number $x$, that is, $f$ is the constant function $f(x) = -2$.
Substituting into the original equation to check, this constant function $f(x) = -2$ also satisfies the condition: Left side $=$ Right side.
Combining the above, there are exactly two functions satisfying the condition: $f(x) = -2$ and $f(x) = x^2 - 4$.

Alternative solution: Let $P(x, y)$ denote the substitution of $(x, y)$ into the original equation.
Suppose $f(a) = f(b)$, then for all $y \in \mathbb{R}$ we have
$$P(a, y), P(b, y) \Rightarrow f(a^2 - y) = f(b^2 - y).$$
If $a^2 - b^2 = t \neq 0$, then $f$ will be a function with period $t$. For all $x \in \mathbb{R}$ we have
$$P(x, y), P(x, y + t) \Rightarrow 4t(f(x) + 2) = 0 \Rightarrow f(x) = -2.$$
Substituting back into the original equation, we know that $f(x) = -2$ is a solution.
If $a^2 - b^2 = 0$, then this means $a = \pm b$. Therefore we can deduce
$$f(a) = f(b) \Rightarrow a = \pm b.$$
On the other hand,
$$P(x, 2) \Rightarrow f(f(x) + 2) = f(x^2 - 2) \Rightarrow f(x) + 2 = \pm(x^2 - 2),$$
that is, for all $x \in \mathbb{R}$ we have
$$f(x) = -x^2 \lor f(x) = x^2 - 4.$$
If there exists a real number $x$ such that $f(x) = -x^2$,
$$P(x, x^2) \Rightarrow f(0) = f(0) + 4(x^2 - 2)(-x^2 + 2) \Rightarrow x = \pm\sqrt{2},$$
that is, $f(x) = x^2 - 4$ holds for all cases except $x = \pm\sqrt{2}$. Also
$$(\pm\sqrt{2})^2 - 4 = -(\pm\sqrt{2})^2 = -2.$$
So $f(x) = x^2 - 4$ actually holds for all $x \in \mathbb{R}$, and substituting back into the original equation we know that $f(x) = x^2 - 4$ is a solution.
In summary,
$$f(x) = -2, \quad f(x) = x^2 - 4$$
are all the solutions to the functional equation.