What is the role of NaHSO3 in syn diol formation from alkenes with OsO4?
After you add the $\ce{OsO_4}$ to make the reaction, you'll need to reduce the remaining $\ce{OsO_4}$ maintaining the $\ce{pH}$ of the solution constant. 

In addition:


  The hydrolysis of the osmate ester proceeds by simple substitution
  reactions. The $\ce{NaHSO3}$, a mild reducing agent, facilitates ligand
  subsitution by reducing the $\ce{Os}$ further.


The Art of Writing Reasonable Organic Reaction Mechanisms, Robert B. Grossman, p 293

And $\ce{NaHSO_3}$ can do both and so you will use it. You also can add $\ce{Na_2SO_3}$.



I found also something interesting on JACS, it's mentioned at least four times in the document that $\ce{NaHCO3}$ is used as buffer. Also there is a lot of information in the references of this article.


  While the reaction is normally run under basic conditions
  ($\ce{K_2CO_3}$, $\ce{pH} = 12.2$, aqueous layer), it is possible to
  buffer the system with $3$ equiv of $\ce{NaHCO3}$ ($\ce{pH} = 10.3$,
  aqueous layer).$^{38, 39}$ Buffering of the reaction mixture does not
  affect the ee, but it can have a beneficial effect on the yield when
  base-sensitive substrates are used or base-sensitive products are
  formed. Thus, the AD of allyl or cinnamyl halides (cf. eqs $1$ and
  $3$) should be performed under buffered conditions to minimize epoxide
  formation (see eq $1$, the yield with allyl bromide as substrate is
  between $61$ and $74%$ under buffered conditions; in the absence of
  $\ce{NaHCO3}$ only $40-50$ % of diol is obtained38). Unfortunately,
  the AD reaction does not turn over if $\ce{K_2CO_3}$ is replaced
  entirely by $\ce{NaHCO3}$.


Hartmuth C. Kolb, Michael S. VanNieuwenhze, K. Barry Sharpless
Chem. Rev., 1994, 94 (8), p 2494