I really like this question because it digs at wanting to understand what's going on beneath the equations you've been given to know why that equation works.
Orthocresl answered it really well, but I might be able to get a little more to what pH and pOH are and give a little more understanding to what's meant by acidic and basic, and what equilibrium is. I feel like that might be as much your question as whether or not H2O can react with CO3 2- (apologies, first time answering a question here, and not sure how to format this correctly).
tl;dr version up front:
Pure water has an equal amount of H+ and OH-, both of them exist in water. We consider this neutral, neither acidic or basic. Anything that moves the concentration to favor more H+ is considered an (Arrhenius) acid, anything that moves the concentration to favor more OH- is considered a base. Even so, there will still be both H+ and OH- in the solution. Individual water molecules will continue to accept protons (H3O+, usually shown as H+) and donate protons (OH-), but there will be more water molecules accepting a proton to form H3O+ than donating them to form OH- at lower pH.
non tl;dr version:
Very, very few chemical reactions are truly one way/irreversible unless the products leave the system entirely. Almost all will be occurring in both directions at the same time. But, one will be occurring more/faster. Let's say reaction A occurs twice as fast as the reverse reaction B (A: a + b -> c; B: c -> a + b). Since A occurs twice as fast a B, then there will be more products, c, than reactants, a and b. However, c will still be producing a and b. Eventually you'll get enough c that the reaction will reach a point that it will look like the reaction stopped. If reaction A occurs twice as fast as reaction B, this will be when there's twice as much products as reactants.
Basically, if a + b produce c twice per minute, and c produces a + b once per minute, you'll eventually get twice as much c as you have a and b, and the reaction will look like it stopped from a macro point of view, but will continue to occur in your solution.
Take what I said there from a conceptual view. It's useless for practical purposes. If you want to know the math, this video explains it really well (https://www.youtube.com/watch?v=2PM1yc_z4Bk&list=PL2ub1_oKCn7qmUZ80MJDPaRTdgJm8ZenX).
I don't see a good way to not discuss different definitions of acids even though I feel it's a little beyond this question. Arrhenius acid is anything that lowers the pH of pure water (base is the opposite). But that's not a good definition. A much better one is the Bronsted definition, anything that donates a proton acts as an acid, and anything that accepts a proton acts as a base (there's a 3rd definition, but it's difficult to understand at first). In this case, your carbonic acid is acting as an acid when it donates a proton to water to form H3O+, and water acts as a base. However, the reverse happens as well. H3O+ will donate a proton to HCO3-. In which case H3O+ is acting as an acid. H2O can also act as a Bronsted acid and donate a proton to form OH-. So yes, CO3 2- can and does act as a base and accept a proton from H2O, and you get OH- floating around.
In any aqueous solution, you'll end up with both H3O+ and OH-. This is usually measured with pH. pH is actually the result of this equation pH = -log(10)([H+]). That equation means that pH is the negative base 10 logarithm of the concentration of protons (hydronium ions) in the solution. However, because we know there's an equilibrium between [H+] and [OH-], we can actually measure this in terms of the concentration of OH- ions instead, pOH. pOH is remarkably, the negative logarithm of the concentration of hydroxide ions, pOH = -log(10)([OH-]).
So let's say you have a pH of 5, a slightly acidic solution. You'll have a pOH of 9. The pOH measures in the opposite direction, pOH below 7 is basic, above 7 is acidic. These two measure will equal 14. A pH of 1 will correspond to a pOH of 13. So even at a very high acidity of pH 1, you will still have H2O acting as an acid and donating protons to form OH-. It will just be happening very slowly, and so there won't be many of them.