What is the justification for these two steps in proof of Baire's theorem?

A locally compact Hausdorff space is regular and so if $x \in O$ is a situation in such a space, where $O$ is open, we can find an open $O_1$ so that $$x \in O_1 \subseteq \overline{O_1} \subseteq O$$ by regularity. After that we can find a relatively compact $U$ so that $x \in U \subseteq O_1$ (because relatively compact sets form a base). It then follows that $\overline{U} \subseteq \overline{O_1} \subseteq O$: i.e. the closure of the relatively compact sets sits inside the original open set (and we can forget about the "auxiliary" $O_1$ again).

So the first part about why $U$ exists for $x' \in U_0-A_1$: it's the previous argument (lemma?) applied to the open set $O=U_0-A_1$ (which is open as an "open minus closed" set) and $x=x'$. The resulting $U$ is then called $U_1$ and is the next step in the recursion.

Next, $A_2$ also has empty interior. As $U_1$ is non-empty (as witnessed by $x'$ from before) it cannot be a subset of $A_2$. So $U_1 - A_2$ is open (open - closed again) and non-empty and "the show can continue" (we can proceed with the recursion).

Hope this answers your doubts.