Assumed I have a one-dimensional function in cartesian coordinates, which is symmetric concerning the y-axis, i.e. for example $f(x)=\cos(x)$. Now the laplace operator for that function is
$$\Delta f(x)=\frac{d^2}{dx^2}f(x)$$
When transforming that into polar coordinates, $f(x)$ becomes $f(r)$, and
$$\Delta_{xx}\Rightarrow\frac{d^2}{dr^2}f(r)+\frac{1}{r}\frac{d}{dr}f(r)$$
while neglecting the angular dependence (i.e. $\frac{d^2}{d\phi^2}=0$).
My problem here is: I should get the same result, regardless of the coordinate system. But the results are not equal, due to the additional term in the radial laplace operator.
Does that thus mean that I simply can not do the transfer written above? Or do I have to add additional parameters?
Siminore is correct. The one-dimensional Laplacian is not the two-dimensional laplacian with one dimension set to $0$.
If we have a function $\hat f(x)$, we can consider it as a function of two variables by the simple device of setting $f(x, y) = \hat f(x)$ for all $y$. Now because $f$ does not change in the $y$ direction, the directional derivative in that direction is $0$. Since this is constant, it does not contribute anything to the divergence, and thus the laplacian for this two-dimensional function is the same as for it's one-dimensional precursor.
But what happens when we convert to polar coordinates. You are assuming that now there is no change in the $\phi$ direction, but this is not the case. $f(x,y)$ is constant with respect to $y$. It is not constant with respect to $\phi: f(r,\phi) = \hat f(r\cos \phi)$. So $\frac {\partial^2 f}{\partial \phi^2} \ne 0$.
If you instead define $f_2(r, \phi) = \hat f(r)$, this is a different function than $f$. For example, if $\hat f(x) = ax$ for some $a$, then $f(x,y)$ is a plane while $f_2(r, \phi)$ is a cone. It is a different function, and so has a different gradient and a different laplacian.