I often see Vega in the Heston model specified as: \begin{align*} \nu & = \frac{\partial C}{\partial v} = \frac{\partial C}{\partial v_0} 2 \sqrt{v_0} \end{align*} where $v = \sqrt{v_0}$.
Why are we setting $v = \sqrt{v_0}$?
Where is the square-root and "$2$" coming from?
Let $V$ denote the variance and $v$ the volatility, i.e. $V=v^2$. The natural argument for the option price under a stochastic volatility model is typically the variance, i.e. $C_\text{SV}=C_\text{SV}(S_0,V_0,...)$. However, using the chain rule, we can compute vega in terms of the volatility: $$\nu=\frac{\partial C_\text{SV}}{\partial v}=\frac{\partial C_\text{SV}}{\partial V}\frac{\partial V}{\partial v}=\frac{\partial C_\text{SV}}{\partial V}\frac{\partial v^2}{\partial v}=\frac{\partial C_\text{SV}}{\partial V}2v=\frac{\partial C_\text{SV}}{\partial V}2\sqrt{V}.$$ We do this in order to resemble the Black-Scholes vega which is the partial derivative of the Black-Scholes option price, $C_\text{BS}=C_\text{BS}(S_0,\sigma,...)$, with respect to $\sigma$. Of course, when we have $\frac{\partial C_\text{BS}}{\partial \sigma}$, we can easily infer $\frac{\partial C_\text{BS}}{\partial \sigma^2}$ using again the chain rule. This would be the Black-Scholes vega with respect to the variance.
The model-free put-call parity implies that European-style put and call options have the same vega (with respect to volatility or variance).
Writing $V_0$ and $v_0=\sqrt{V_0}$ emphasises that we talk about the spot variance and volatility.
