I worked through this conversion and it all makes sense except for one small part. Shouldn't $(q_1q_2)$ go to $q_1$ in the DFA on input $0$, not a self loop?
We have state $q_iq_2$ to begin with because $q_0$ goes to $q_1$ and $q_2$ on a $1$.
Then trying to figure out what states $(q_1q_2)$ goes to:
On a $0$ $(q_1q_2)$ goes to $q_1$ because in the original NFA $q_1$ goes to $q_1$ and $q_2$ goes to the empty set. On a $1$ $(q_1q_2)$ goes to $q_3$ because in the original NFA $q_1$ goes to $q_3$ and $q_2$ goes to the $q_3$.
If I'm not mistaken, the technique is: for each transition $q \to^a Q'$ of the NFA (NFA transitions are from states to sets of states), you take $$q \to^a \text{state}\big(Q'\cup\{q'' ~|~q'\in Q'~\text{and}~q'\to^\epsilon Q''~\text{and}~q''\in Q''\}\big)$$
This is just a formal way to say "in $q\to^a Q'$, take into account also the $\epsilon$-transitions from $Q'$".
Here, it's $q_1 \to^0\{q_1\}$. Therefore, this results to $q_1 \to^0 \text{state}\big(\{q_1\}\cup\{q_2\}\big) = (q_1, q_2)$.
The function "state" is just to map a set of states of the initial NFA to the corresponding state of the DFA.