The set of differentiability will depend on the enumeration. Indeed, we can make $h$ differentiable at a given irrational point $\alpha$ by ensuring that $|r(k)-\alpha|\ge \frac{1}{k} \quad \text{ for all }k$ This requirement slightly restricts our choice of $r(k)$ but still allows for enumeration of all rationals. To see that $h'(\alpha)=0$, notice that $h(\alpha+1/k)-h(\alpha)\le \sum_{j=k}^\infty 2^{-j}=2^{1-k}$ is much smaller than $1/k$, and similarly for $h(\alpha)-h(\alpha-1/k)$.
On the other hand, we can also make $h$ non-differentiable at a given irrational point $\alpha$ by choosing $|r(k)-\alpha|<4^{-k}$ when $k$ is even, and using the odd indices to enumerate the rest of rationals. With this enumeration $h(\alpha+4^{-k})-h(\alpha)>2^{-k}$, which is much larger than $4^{-k}$. Hence $\limsup_{x\to \alpha}\frac{h(x)-h(\alpha)}{x-\alpha}=\infty$.