A certain kind of non-linear mapping from $\mathbb{R}^2$ to $\mathbb{R}^2$

$$T(x_1,x_2) = \big((x_1^2x_2)^{1/3},x_1+x_2\big)$$

Based on Doug M's strategy, here's a symmetrical one:

$$T(x,y) = \left((x^3+y^3)^\frac{1}{3},(x^3+y^3)^\frac{1}{3}\right)$$

Or, with the same idea, an even simpler one:

$$T(x,y) = \left((x^3+y^3)^\frac{1}{3}\,,\,0\right)$$

Still with the same idea, but dispensing with symmetry, here's a bijective one:

$$T(x,y) = \left((x^3+y^3)^\frac{1}{3},(x^3-y^3)^\frac{1}{3}\right)$$

What about

$f(x,y) = \begin{cases} (0,0) & x \ne 0 \\ (x,y) & x = 0\end{cases}\\$

Conceptually, the given condition requires that if we restrict $f$ to any line through the origin, it is linear. Since each of those lines only overlaps at the origin, we can construct $f$ easily by constructing its value on each of those lines one by one.

In this case, we choose $f$ to be zero on every line except on the y axis. On the y axis, it just looks like a the identity function. The function constructed this way is not linear because its value on the y axis doesn't "agree" with its value everywhere else. If we try to check whether the law of linearity applies, we check two points $p_1, p_2$, and verify that

$f(p_1 + p_2) = f(p_1) + f(p_2)$

Now, if both $p_1$ and $p_2$ are on the y axis, or both of them are not on the y axis, this equality will hold. However, if one of them is on the y axis and the other isn't, you'll run into problems. For example:

$p_1 = (1,0)$ $p_2 = (1,1)$ $f(p_1+p_2) = f(2,1) = (0,0)$ $f(p_1) + f(p_2) = (1,0) + (0,0) = (1,0)$

In general, you can see that there are infinitely many ways that we could choose different linear functions on all the lines through the origin and "piece together" a novel f that violates the constraints, even a continuous or differentiable one, as other people have demonstrated.