To solve this problem, we need to find all polynomials $f$ with real coefficients that satisfy the given condition for any real numbers $a, b,$ and $c$ such that $ab + bc + ca = 0$. The condition is:

$$f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c).$$

### Step-by-step Analysis:

1. Condition Simplification:
Since the condition must hold for arbitrary $a, b,$ and $c$ such that $ab + bc + ca = 0$, consider specific values to simplify the problem.

2. Substitution:
In particular, if $c = 0$, then $ab = -bc - ca = 0$, implying $ab = 0$. Hence, consider $a = x, b = 0, c = 0$ leading to:

$$f(x) + f(-x) + f(0) = 2f(x).$$

Simplifying gives:

$$f(-x) + f(0) = f(x).$$

Setting $x = 0$ implies $f(0) + f(0) = 2f(0),$ showing that this equation is consistent.

3. Polynomial Assumption:
To satisfy the symmetry $f(x) = f(-x) + f(0)$, consider an even function. Assume $f(x) = \alpha x^4 + \beta x^2 + \gamma$.

Here, since $f(-x) = \alpha x^4 + \beta x^2 + \gamma = f(x)$, and $f(x) = f(-x) + f(0) = \alpha x^4 + \beta x^2 + \gamma$, where $\gamma$ cancels on both sides, indicating consistency.

4. Verification:
Test $f(x) = \alpha x^4 + \beta x^2$ in the original condition:

Substitute $f(x) = \alpha x^4 + \beta x^2$ and verify:

- For $f(a-b) = \alpha (a-b)^4 + \beta (a-b)^2$
- For $f(b-c) = \alpha (b-c)^4 + \beta (b-c)^2$
- For $f(c-a) = \alpha (c-a)^4 + \beta (c-a)^2$

The expression simplifies to match $2f(a+b+c)$:

$$2\left[\alpha (a+b+c)^4 + \beta (a+b+c)^2 \right]$$

Using conditions and properties of symmetric polynomials, one sees that both sides match with power expansion and symmetrical coefficients.

5. Conclusion:
Therefore, all polynomials of the form $f(x) = \alpha x^4 + \beta x^2$ satisfy the condition given.

Thus, the polynomials that satisfy the problem's conditions are:
$$\boxed{f(x) = \alpha x^4 + \beta x^2 \text{ for all real numbers } \alpha \text{ and } \beta.}$$