Without loss of generality, we may assume that $f(0) = 0$. Let $y > 0$ and $n$ a positive integer. For $x = n y$ we get
$$2 f(n y) = f((n + 1) y) + f((n + 2) y).$$
The sequence $a_n = f(n y), n = 0, 1, 2, \ldots$, satisfies the second order linear recursive relation
$$a_{n+2} = -a_{n+1} + 2 a_n$$
with $a_0 = f(0) = 0, a_1 = f(y)$. The characteristic equation is $t^2 + t - 2 = 0$, having the roots $t_1 = 1, t_2 = -2$. It follows that $a_n = \alpha + \beta (-2)^n$, where $\alpha + \beta = 0$ and $\alpha - 2 \beta = f(y)$. We get
$$f(n y) = a_n = \frac{1 - (-2)^n}{3} \cdot f(y)$$
In particular, we have $f(4 y) = -5 f(y)$. On the other hand, from (3), we obtain $f(2 y) = -f(y)$, hence $f(4 y) = -f(2 y) = f(y)$, for any $y \geq 0$. That is $f(y) = 0$ for any $y \geq 0$.
Let $x \in \mathbb{R}$. Then
$$2 f(x) = f(x + |x|) + f(x + 2|x|) = 0$$
hence $f(x) = 0$. The desired functions are all constant functions.

A simple checking shows that any constant function is solution to the functional equation.

If $x_1, x_2 \in \mathbb{R}, x_1 > x_2$, we can find $x \in \mathbb{R}$ and $y \geq 0$ such that $x_1 = x + 4y$ and $x_2 = x + y$. Indeed, solving the system in $x$ and $y$, we get
$$x = \frac{1}{3}(4 x_2 - x_1), \quad y = \frac{1}{3}(x_1 - x_2)$$
Replacing $y$ by $2y$ in the functional equation we obtain
$$2 f(x) = f(x + 2y) + f(x + 4y),$$
hence we get
$$f(x + y) + f(x + 2y) = f(x + 2y) + f(x + 4y)$$
that is
$$f(x + y) = f(x + 4y), \quad x \in \mathbb{R}, y \geq 0$$
From (1) it follows $f(x_1) = f(x_2)$, for any $x_1, x_2 \in \mathbb{R}$, hence $f$ is constant function.

A simple checking shows that any constant function is solution to the functional equation.

As in the previous solution, we have
$$2 f(x) = f(x + y) + f(x + 2y) = f(x + 2y) + f(x + 4y),$$
hence, we get relation (1). Taking $x = -y, y \geq 0$, from (1) we obtain $f(0) = f(3y), y \geq 0$, that is $f$ is constant on $[0, \infty)$. Taking $x = -3y$, it follows
$$f(-3y) = f(y) = 0, \quad y \geq 0$$
hence $f$ is constant on $(-\infty, 0]$. Therefore, all solutions are given by the constant functions.

A simple checking shows that any constant function is solution to the functional equation.