Why is this method wrong on calculating this multivariable limit?

Your very first statement

In particular, in $\Bbb{R}^2$ we have polar coordinates to switch on and we have: $\lim_{{(x,y)}\to(0,0)}f(x,y) = L$ iff $\lim_{r\to0^+}f(r\cos(\theta),r\sin(\theta)) = L$

is not very meaningful yet, because you haven't put a quantifier over $\theta$. I guess you meant the following:

In particular, in $\Bbb{R}^2$ we have polar coordinates to switch on and we have: $\lim\limits_{{(x,y)}\to(0,0)}f(x,y) = L$ iff for every $\theta\in \Bbb{R}$, $\lim\limits_{r\to0^+}f(r\cos(\theta),r\sin(\theta)) = L$.

Even if this is what you meant, it is false, and this is a very common misconception (unfortunately there are several notes which promote the use of polar coordinates for solving limits, without carefully explaining the subtleties).

The implication $\implies$ is true, while the reverse implication is false. This is because if you fix a value of $\theta$, then $\lim_{r\to0^+}f(r\cos(\theta),r\sin(\theta))$ is taking a limit of a single-variable function along a certain straight half-line (i.e it is a one-sided limit along a straight line) which is clearly a much weaker condition than what is actually required ($\lim_{(x,y)\to (0,0)}f(x,y)$ requires the limit to exist regardless of how you approach the origin: straight line, curvy line, zig-zag/criss-cross/oscillatory, whatever).

In fact your function is a perfect example, because it shows that along EVERY straight line to the origin, the limit of the function is $0$, yet despite this the multivariable limit $\lim_{(x,y)\to (0,0)}f(x,y)$ does not exist.

Just to drive the point home, let's write out in terms of quantifiers what each statement means:

1. $\lim_{(x,y)\to (0,0)}f(x,y) = L$ means:

for every $\epsilon>0$, there is a $\delta>0$ such that for all $(x,y) \in \Bbb{R}^2$, if $0< \lVert (x,y)\rVert<\delta$ then $|f(x,y) - L| < \epsilon$.

2. For every $\theta\in \Bbb{R}$, $\lim_{r\to 0^+} f(r\cos\theta, r \sin \theta) = L$ means:

For every $\theta\in \Bbb{R}$, for every $\epsilon>0$, there is a $\delta>0$ such that for all $r>0$, if $0<r<\delta$ then $|f(r\cos\theta, r \sin \theta) - L|< \epsilon$.

3. $\lim_{r\to 0^+}f(r\cos\theta, r \sin\theta) = L$, uniformly in $\theta$, means:

For every $\epsilon>0$ there is a $\delta>0$ such that for all $r>0$ and $\theta\in \Bbb{R}$, if $0<r<\delta$ then $|f(r\cos\theta, r \sin \theta) - L|< \epsilon$.

Note the differences in the statements, especially between 2 and 3 in terms of the quantifiers. We have $(1)\iff (3)$, and $(1)\implies (2)$ (so trivially $(3)\implies (2)$) but your specifc function shows that $(2)\nRightarrow (1)$.

In (1) and (3), the $\delta$ depends only on $\epsilon$, while in $(2)$, the $\delta$ depends on $\theta$ and $\epsilon$ (which is why order of quantifiers matters). Also, if you've seen the concept of uniform continuity, then you'll observe that it is a similar switch in the order of quantifiers which distingusihes between $(2)$ and $(3)$.