How do I calculate a sum $\sqrt{\frac {\hbar{(v)} }{4G}} r$ with respect to a changing $v$?

Both the square root function and the integration operation handle multiplicative constants nicely. In particular, if you have non-negative real numbers $a,b$ then the identity $\sqrt{ab}=\sqrt a\sqrt b$ holds. So for your title quantity we can say $$\sqrt{\frac{\hbar (v)}{4G}}r=r\sqrt{\frac{\hbar}{4G}}\sqrt v=C_0\sqrt v$$ where we will use $C_0$ to refer to that constant portion moving forward. Then we can write the listed integral as $$\int_{v_i}^{v_f}C_0\sqrt v\text dv=\frac 23C_0\left(v^{\frac 32}\right|_{v_i}^{v_f}$$ since a multiplicative constant may be moved outside an integral. Evaluation of the integral result can be addressed separately.

Edit: from the new listed sum of terms, the individual sum term would be expressed more like

$$\sqrt{\frac{\hbar v}{4Gv_f}}r$$

and the proposed sum can be approached with

$$\sqrt{\frac{\hbar}{4Gv_f}}r\sum_{v=1}^{v_f}\sqrt v$$

but this has a similar problem to the original in that if $v_f\to\infty$ then the result of the sum will approach infinity as well. See this question Conversion of Riemann Sum to Integral with Square Root for a very similar problem and how much of a $dx$ is required to make the sum converge...

Edit 2, corrected:

The area under a square root curve $f(x)=c\sqrt x$ in an interval $[a,b]$ is always $\frac 23c(b^{\frac 32}-a^{\frac 32})$. Summing over a large set of consecutive integers plugged in for $x$ in this function will closely approximate this result, especially as the number of values increases. So for the listed value of $v_f$ we should expect that the result of the listed sum will be within the best possible error tolerances of $\frac 23C_0v_f=\frac 23v_fr\sqrt{\frac{\hbar}{4G}}$.