I don't have any intuition in judging how fast a term of the order $\mathcal{O}( h^2 |\log(h)|)$ is decreasing as $h \to 0$, so i tried comparing it with terms of the form $\mathcal{O}( h^\alpha )$ and simply compute:
In Mathematica, for the above plot:
Needs["PlotLegends`"]; logterm = Table[(1/n)^2*Abs[Log[1/n]],{n,1,1000}] ; table2 = Table[(1/n)^(2),{n,1,1000}]; table3 = Table[(1/n)^(1.5),{n,1,1000}]; table4 = Table[(1/n)^(1.82),{n,1,1000}]; ListLinePlot[{logterm,table2, table4, table3},PlotLegend->{"h^2 |log(h)|", "h^2", "h^1.82","h^1.5"},LegendPosition->{0.45,0.0}, LegendSize->0.6, PlotStyle->{Blue, Red,Green,Gray}]
First it seems, that $\mathcal{O}( h^2 |log(h)|)$ is slower than $\mathcal{O}( h^{1.82} )$ and faster than $\mathcal{O}( h^{1.5} )$. But if you evaluate it for extremely small $h$, you'll notice that $\mathcal{O}( h^{1.82} )$ is indeed slower:
I assume stuff like this happens for all $0 < \alpha < 2$, even if it's not possible to find a small enough $h$ in my computer anymore because of the memory restrictions.
Can anyone confirm or give some explanations, why it's right to assume, that i can think of an $\mathcal{O}( h^2 |\log(h)|)$-term as something slower than $\mathcal{O}( h^{2} )$, but faster than $\mathcal{O}( h^{\alpha} )$ for all $0 < \alpha < 2$ as $h \to 0$?
Yes, you are exactly right. The reason is that for every $\epsilon>0$, $$\lim_{h\to 0} h^\epsilon \log h = 0$$ which follows from L'Hospital's rule: $$\lim_{h\to 0}\frac{\log h}{h^{-\epsilon}} =\lim_{h\to 0}\frac{1/h}{-\epsilon h^{-\epsilon-1} } = \lim_{h\to 0} h^{\epsilon}/(-\epsilon ) =0$$
Therefore, for every $\alpha<2$ $$ \lim_{h\to 0}\frac{h^2\log h}{h^\alpha} = 0 $$ which together with $$ \lim_{h\to 0}\frac{h^2}{h^2\log h} = 0 $$ confirms your conjecture.
Nothing special about $2$ here. Logarithmic factors are almost negligible compared to powers of $h$. (Although in borderline cases, like the convergence of $\int_{0}^1 \frac{1}{h\log^p h}$, they can be decisive.)