The idea behind this is that a weak acid like $CH_3CO_2H$ will break apart, but won't break apart completely. If you have some $CH_3CO_2H$ in water, most of it will remain as $CH_3CO_2H$, so in the net ionic equation it is written whole, but some will separate into $CH_3CO_2^-$ and $H^+$ ions. This relationship is quantified by the $K_a$ of the acid. Essentially, the dissociation reaction $CH_3CO_2H \rightarrow CH_3CO_2^-+H^+$goes both ways: $CH_3CO_2H$ decomposes into $CH_3CO_2^-$ and $H^+$, and $CH_3CO_2^-$ reacts with $H^+$ for form $CH_3CO_2H$. Since the forward reaction happens at a rate proportional to the concentration of $CH_3CO_2H$, and the backwards reaction happens at a rate proportional to the product of the concentrations of $CH_3CO_2^-$ and $H^+$, eventually the ratio $\frac{[CH_3CO_2^-][H^+]}{[CH_3CO_2H]}$ becomes constant, and we call this constant the $K_a$ of the $CH_3CO_2H$. For $CH_3CO_2H, K_a=1.7\times 10^{-5}$.

Now its time to connect all this back to your question. When you add $Ba(OH)_2$ to a solution of $CH_3CO_2H$, the $OH^-$ ions present from the dissociation of $Ba(OH)_2$ react with the $H^+$ ions from the (partial) dissociation of $CH_3CO_2H$. This lowers the concentration of $H^+$, and so more $CH_3CO_2H$ dissociates to keep the ratio of concentrations constant at the $K_a$. Therefore, in the presence of enough $Ba(OH)_2$, $CH_3CO_2H$ dissociates almost completely, even though, on its own, $CH_3CO_2H$ dissociates only to a very small extent.