First we write $(4x + g(x)^2)g(y) = 4g(\frac{y}{2} \cdot g(x)) + 4xy \cdot g(x)$ as
$$\left(x + \left(\frac{g(x)}{2}\right)^2\right) \frac{g(y)}{2} = \frac{1}{2} g\left(y \cdot \frac{g(x)}{2}\right) + xy \cdot \frac{g(x)}{2}.$$
Let $f(x) = g(x)/2$, then the original equation can be transformed into
$$(x + f(x)^2)f(y) = f(yf(x)) + xyf(x). \quad (1)$$
Substituting $x = 1$ into (1), we obtain
$$(1 + f(1)^2)f(y) = f(yf(1)) + yf(1). \quad (2)$$
Substituting $y$ with $1, f(1), f(1)^2$ respectively into (1), we can obtain
$$f(f(1)) = f(1)^3, \quad (3)$$
$$f(1)^3 + f(1)^5 = f(f(1)^2) + f(1)^2, \quad (4)$$
$$f(1)^7 + 2f(1)^5 - f(1)^4 - f(1)^2 = f(f(1)^3). \quad (5)$$
Substituting $y = 1$ and $x = f(1)$ into (1), we can obtain
$$f(1)^2 + f(1)^7 = f(f(1)^3) + f(1)^4. \quad (6)$$
From (5) and (6) we derive $f(1)^5 = f(1)^2$. Therefore $f(1) = 0$ or $f(1) = 1$.
If $f(1) = 0$, then from (2) we know that for all $y$ we always have $f(y) = f(0)$, so we can obtain that $f$ is a constant function, therefore the unique function satisfying this case is the zero function. If $f(1) = 1$, then from (2) we know that for all $y$ we always have $f(y) = y$, and we verify that this function satisfies (1). So the solutions of (1) are $f(x) = 0$ and $f(x) = x$. Since $g(x) = 2f(x)$, the solutions of the original equation are $g(x) = 0$ and $g(x) = 2x$.