No sign flipping while figuring out the emf of voltaic cell?
Take a look at the two half reactions:

$$
\begin{align}
\ce{Ag+(aq) + e- &amp;→ Ag(s)} &amp;\qquad E^\circ &amp;= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &amp;→ Sn(s)} &amp;\qquad E^\circ &amp;= \pu{-0.14 V}
\end{align}
$$

If there is an electron for grabs (like the ones in the wire of a voltaic cell), $\ce{Ag+(aq)}$ and $\ce{Sn^2+(aq)}$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.


  No sign flipping while figuring out the emf of voltaic cell?


Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.

$$E_\text{cell}^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ$$

If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)