Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function satisfying the following conditions:
1. $f(0,x)$ is non-decreasing.
2. For any $x, y \in \mathbb{R}$, $f(x,y) = f(y,x)$.
3. For any $x, y, z \in \mathbb{R}$, $(f(x,y) - f(y,z))(f(y,z) - f(z,x))(f(z,x) - f(x,y)) = 0$.
4. For any $x, y, a \in \mathbb{R}$, $f(x+a, y+a) = f(x,y) + a$.

We aim to find all such functions $f$.

First, define $h(x) = f(0,x)$. Given that $h(x)$ is non-decreasing, we can use the functional equations to derive the form of $f$.

From condition 4, substituting $x$ and $y$ with $x-a$ and $y-a$ respectively, we get:
$$f(x,y) = h(y-x) + x.$$

Using condition 2, $f(x,y) = f(y,x)$, we have:
$$h(y-x) + x = h(x-y) + y.$$
This implies:
$$h(-x) = h(x) - x.$$

Next, we use condition 3. Consider $P_3(0,x,2x)$:
$$(h(x) + x - h(2x))(h(2x) - h(x)) = 0.$$
This implies that for all $x \neq 0$, either $h(2x) = h(x)$ or $h(2x) = h(x) + x$.

### Case 1: $\exists u > 0$ such that $h(2u) = h(u)$
If $h(2u) = h(u)$, then $h(x)$ is constant for $x \geq u$. By extending this argument, we find that $h(x) = a$ for some constant $a$ and for all $x > 0$. For $x < 0$, using the non-decreasing property and symmetry, we get $h(x) = a + x$. Hence, $f(x,y) = a + \min(x,y)$.

### Case 2: $h(2x) \neq h(x)$ for all $x > 0$
If $h(2x) = h(x) + x$ for all $x > 0$, then $h(x)$ must be of the form $h(x) = a + x$ for $x > 0$. For $x < 0$, using the non-decreasing property and symmetry, we get $h(x) = a$. Hence, $f(x,y) = a + \max(x,y)$.

Thus, the functions $f$ that satisfy all given conditions are:
$$\boxed{f(x,y) = a + \min(x,y) \quad \text{or} \quad f(x,y) = a + \max(x,y) \quad \text{for any } a \in \mathbb{R}.}$$