About combining a cation and an anion
I really like your process of trying to work it out! Trying to find patterns is a great way to look at this (and all science really). I think all you're really missing is some nomenclature, as this appears to be primarily a question of understanding of what's actually going on.

The 3+ alumnium ion ($\ce{Al^{3+}}$) and 2- oxygen ion ($\ce{O^{2-}}$) form a crystal lattice in solid form.

(A crystal lattice is a $3D$ arrangement of ions in a repeating pattern, in which each ion is interacting with more than one other ion)

The total charges need to balance out per "unit cell" (an intermediate-level chemistry concept that means the "smallest repeating subunit of a crystal lattice.")
So this means that you need to balance the total charges by changing the number of ($\ce{Al^{3+}}$) and ($\ce{O^{2-}}$) ions to get the correct chemical formula.

In this case: $(3+) *(2) = (6+)$ balances $(2-) *(3) = (6-)$, giving $\ce{2 Al^{3+}}$ and $\ce{3 O^{2-}}$.

Ionic compounds that have an overall neutral charge are written without the charge superscripts,  thus, $\ce{Al^{3+}_2O^{2-}_3}$ is written as $\ce{Al_2O_3}$ 

If you're actually looking at precipitation reactions (where the solid salt forms from a solution), you have to balance the equations.

Thus to form $\ce{Al_2O_3}$ the correct ratio would be: $\ce{2 Al^{3+} + 3 O^{2-}} -&gt; Al_2O_3$

Similarly for $\ce{CaBr_2}$, the correct ratio would be: $\ce{Ca^{2+} + 2 Br^-} -&gt; CaBr_2$.

For $\ce{Na_2CO_3}$ you're applying the exact same concept, except you're dealing with an anionic (negative charged) molecule instead of an atom.

Thus, to form $\ce{Na_2CO_3}$ the correct ratio would be: $\ce{2 Na^+ + CO^{2-}_3} -&gt; Na_2CO_3$.