What you are missing is understanding of general nature of redox reactions. Redox reactions need two half reactions to complete. You can't arbitrarily select these two half reactions. They has to be chosen as instructions given in the problem. For example, for your given problem, it should be noted the medium of the reaction, whether it is acidic or basic or neutral.
Suppose the question asked is: Balance the following redox equation in acidic medium.
$$\ce{I- (aq) + MnO4- (aq) -> MnO2 (s) + I2 (aq)}$$
To solve this, you can directly get two half reactions from a textbook or balance it old-fashioned. The oxidation half reaction is:
$$\ce{2I- <=> I2 }$$
It is already balanced by mass, so add electrons (no mass) to balance the negative charge:
$$\ce{2I- <=> I2 + 2e-}$$
Now oxidation half reaction is completed. The reduction half reaction is:
$$\ce{MnO4- <=> MnO2 }$$
However, it was stated that reaction is in acid medium. Therefore we should add acid to reactants' side:
$$\ce{MnO4- + H+ <=> MnO2 }$$
Now, we have to balance mass first. Since the reaction is in acidic aqueous medium we have plenty of $\ce{H2O}$ molecules around to balance oxigen as well as hydrogen molecules (by $\ce{H+}$ ions as needed):
$$\ce{MnO4- + 4H+ <=> MnO2 + 2H2O }$$
Now equation is balanced by mass. Doing so, we increase net positive charges in left hand side by 3, so we should neutralize it by adding 3 electrons to left hand side because right hand side is neutral:
$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O }$$
Now reduction half reaction is also completed. To achieve balanced redox reaction, simply add balanced oxidation and reduction half reactions in order to cancel unwanted electrons:
$$\ce{2MnO4- + 8H+ + 6I- -> 2MnO2 + 3I2 + 4H2O }$$
This redox reation is forward reaction because it has a net positive potential (refer reduction potentials of two half reactions). The major point is the reaction starts in acidic medium and stay acidic at the end. Keep in mind that if you have used very strong acidic conditions, $\ce{MnO2}$ would be further reduced to $\ce{Mn^2+}$ according to the half reaction: $\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O }$. But oxidation half reaction would stay unaffected, yet final redox reaction would be:
$$\ce{2MnO4- + 16H+ + 10I- -> 2Mn^2+ + 5I2 + 8H2O }$$
Main point is changing the conditions most certainly change the reaction. We cannot arbitrarily choose it. For example, to balance the given equation, we have to use what's available for us. We cannot use what is not available or would be available in near future as a product (e.g., as $\ce{OH-}$ in the example in oxidation reaction in neutral medium you are considering).
For the shake of the argument, let's consider that reaction:
Suppose the question asked is: Balance the following redox equation in neutral medium.
$$\ce{I- (aq) + MnO4- (aq) -> MnO2 (s) + I2 (aq)}$$
Again, to solve this problem, you can directly get two half reactions from a textbook or balance it old-fashioned as described above. The oxidation half reaction does not changed by the conditions and following is the completed equation:
$$\ce{2I- <=> I2 + 2e-}$$
The reduction half reaction is:
$$\ce{MnO4- <=> MnO2 }$$
However, it was stated that reaction is in a neutral medium (the product $\ce{MnO2}$ is possible both in acidic or neutral medium). Therefore we cannot add acid or base to reactants' side but water in either side. Therefore, we can balance loss of oxygen in product side by using water because it is in aqueous medium and we have plenty of $\ce{H2O}$ molecules around:
$$\ce{MnO4- <=> MnO2 + 2H2O}$$
Now we have extra hydrogen in product side, so we don't have a choice but balance that by $\ce{H+}$ in reactant side for now:
$$\ce{MnO4- + 4 H+ <=> MnO2 + 2H2O}$$
It is fact that acid-base neutralization reaction produces $\ce{H2O}$ predominantly. Thus, we can add $\ce{OH-}$ to both sides of the reaction so that it can keep the mass balance of the reaction. Since the reaction is in aqueous medium this won't affect the reaction:
$$\ce{MnO4- + 4H+ + 4OH- <=> MnO2 + 2H2O + 4OH-}$$
$$\ce{MnO4- + 2H2O <=> MnO2 + 4OH-}$$
Now the equation is balanced by mass. However, we have increased net negative charges in right hand side by 3, so we should neutralize it by adding 3 electrons to left hand side to cancel the charges:
$$\ce{MnO4- + 2H2O + 3e- <=> MnO2 + 4OH-} \tag{1}$$
Now reduction half reaction is also completed. To achieve balanced redox reaction, simply add balanced oxidation and reduction half reactions in order to cancel unwanted electrons:
$$\ce{2MnO4- + 4H2O + 6I- -> 2MnO2 + 3I2 + 8OH- }$$
The major point is this reaction starts in neutral medium, but become basic at the end. Again, we didn't have choice to keep it neutral because that't the way this happens in nature. Keep in mind that we did not add anything unavailable during balancing the equation. If you check literature for reduction half reaction of $\ce{MnO4-}$ in neutral medium, you would get nothing but equation $(1)$ as we derived.
