First of all, is there any simple explanation as to why I can't just "divide the expected length with the expected speed"?
But anyway, I have $f(x)=\frac{1}{2}(x-4), x\in \Omega = [4,6]$ for the length
and $g(x)=\frac{1}{6}y^2, y\in \Omega = [2,4]$ for the speed.
Initially I tried calculating the expected time for the journey by dividing the expected length by the expected speed, but clearly that is wrong. How are probabilities like this calculated (and what is the logic behind them)?
If $L$ denotes length and $S$ denotes speed then what you are after is the expectation of $L/S=LS^{-1}$. If $L$ and $S$ are independent then so are $L$ and $S^{-1}$ so that: $$\mathbb EL/S=\mathbb ELS^{-1}=\mathbb EL\mathbb ES^{-1}$$ In general it is not true that $\mathbb ES^{-1}=(\mathbb ES)^{-1}$ so we do not end up with $\frac{\mathbb EL}{\mathbb ES}$.
You can find $\mathbb EL$ and $\mathbb ES^{-1}$ by finding the integrals:$$\int^6_4xf(x)dx\text{ and }\int^4_2y^{-1}g(y)dy$$
Do not forget though that this is correct under condition that $L$ and $S$ are independent.