The answer is any constant function $f$.
Clearly, constant functions are solutions. In the following, we show that there is no solution if $f$ is not a constant function.
Label the equation
$$f(f(x) + y) + f(x + f(y)) = 2f(xf(y)). \qquad (1)$$
Swapping $x$ and $y$, we get $f(f(y) + x) + f(y + f(x)) = 2f(yf(x))$. Comparing with (1), we get
$$f(xf(y)) = f(yf(x)). \qquad (2)$$
Putting $x = 0$ in (2), we find that $f(0) = f(yf(0))$. If $f(0) \neq 0$, then $yf(0)$ runs through all real values. This implies $f$ is a constant function, which is a contradiction. So we have $f(0) = 0$. Putting $x = 1$ in (2), we obtain
$$f(f(y)) = f(yf(1)). \qquad (3)$$
Next, we put $x = 0$ in (1) to get
$$f(y) + f(f(y)) = 0. \qquad (4)$$
Replacing $x$ by $f(x)$ in (1), and using (4), we find that
$$f(-f(x) + y) + f(f(x) + f(y)) = 2f(f(x)f(y)).$$

Swapping $x$ and $y$, and comparing with this equation again, we get
$$f(-f(x) + y) = f(-f(y) + x). \tag{5}$$
Putting $y = f(x)$ in (5) and using (4), we have $f(f(x) + x) = 0$. Then we put $x = y = 1$ in (1) to obtain $f(f(1)) = 0$. It follows from (4) by putting $y = 1$ that $f(1) = 0$. Thus, (3) becomes $f(f(y)) = 0$. Together with (4), we find that $f(y) = 0$ for any $y \in \mathbb{R}$. This is a contradiction as we have assumed $f$ is non-constant.
Therefore, the only solutions are the constant functions.