My lab partner performed one section of a lab on her own while I worked on something else. This section involved first adding a drop of 6 M NaOH to Zn(NO3)2. A precipitate forms. This is repeated in three separate test tubes. In each test tube, a different substance is added. To one, HCl is added. My partner noted that upon addition of 4 drops of HCl, all the precipitate is gone. To the next, NaOH is added. She again notes that all the precipitate dissolves. Finally, she added NH3 and notes that all the precipitate dissolved.

Why would the precipitate dissolve on addition of a base? Wouldn't the concentration of OH- increase, thus resulting in MORE precipitate being formed? I'm just wondering if my understanding is skewed or if the results are.

No, your understanding is not skewed, there is just more chemistry to it than that.
Addition of NaOH causes zinc hydroxide to ppt.
Zn^+2 + 2OH^- ==> Zn(OH)2

Addition of HCl simply neutralizes the NaOH so the Zn(OH)2 ppt dissolves.

Addition of more NaOH (excess OH^-)
Zn^+2 + 2 OH^- ==> Zn(OH)2
more OH^- gives
Zn(OH)2 + 2OH^- ==> Zn(OH)4^-2, a complex ion. Zn is an amphotere meaning that is may act as both a metal and a non-metal. It reacts as a metal to form the hydroxide, then the complex ion is formed and it is part of the anion (that would be Na2Zn(OH)4 although that form of it usually is written as Na2ZnO2 which is sodium zincate.

Addition of NH3 causes the Zn(NH3)4^+2 to form. The NH3 is a stronger complexing agent than hydroxide.

If a complex ion resulted from the addition of NaOH or NH3 [or anything else], would that mean that the solubility of Zn(OH)2 increased? Or does it really have no correlation to the solubility at all?

In a sense I guess you could say the solubility of Zn(OH)2 was increased but it is doing so simply because something else was formed. It is another case of shifting the equilibrium.
Zn^+2 + 2OH^- ==> Zn(OH)2
Zn^+2 + 2NH3 ==> Zn(NH3)4^+2 OR
Zn(OH)2 + 2NH3 ==> Zn(NH3)4^+2 + 2OH^-
Addition of NH3 moves the equilibrium to the right which means more of the NH3 complex is formed and more of the hydroxide dissolves. Addition of more OH^- forces the equilibrium back to the left and that means Zn(OH)2 forms again and there is less of the NH3 complex. The same thing happens with Ag^+ + Cl^- ==> AgCl(s). Now if we add NH3,
AgCl + 2NH3 ==> Ag(NH3)2Cl(aq)==> Ag(NH3)2^+2 + 2Cl^- and all of the AgCl dissolves. Then addition of dilute HNO3 neutralizes the NH3 (decreases NH3) so the equilibrium shifts back to the left and the ppt of AgCl forms again. That is a qualitative test for Ag^+ or Cl^-

In summary, the reason why the precipitate dissolves upon addition of a base is because it forms a complex ion with the metal cation, which increases the solubility of the precipitate. In the case of Zn(OH)2, addition of NaOH or NH3 forms complex ions (Na2ZnO2 or Zn(NH3)4^+2) that have higher solubility than the original precipitate. This is due to the shift in equilibrium caused by the formation of the complex ions. When more base (OH^-) is added, the equilibrium shifts to the right and more complex ions are formed, increasing the solubility. Conversely, when the concentration of OH^- is decreased or neutralized, the equilibrium shifts to the left and the original precipitate reforms.

In conclusion, the addition of a base such as NaOH or NH3 can dissolve the precipitate formed by adding 6 M NaOH to Zn(NO3)2. This is because the base can form complex ions with Zn^+2, which shifts the equilibrium and leads to the dissolution of the precipitate. The solubility of Zn(OH)2 is increased as a result of the formation of these complex ions.