The only error here is the second integral on the RHS should have the dot product as the integrand:
$$ \int_{\Omega} \vec{v} \cdot \, \nabla f \; dV, $$
since $\nabla \cdot (f \,\vec{v}) = f \, \nabla \cdot \vec{v} + \nabla f \cdot \vec{v}$.
Aside from appearance, given an integrable function $f:\Omega \subset \mathbb{R}^3 \to \mathbb{R}$ there is no difference between
$$\int_\Omega f, \quad \int_\Omega f \, dV, \quad\text{and } \quad \int\int\int_\Omega f \, dV$$
There is difference between a multiple Riemann integral and an iterated Riemann integral, for example,
$$\int_{[0,1]^3} f \quad \text{and} \quad\int_0^1 \left(\int_0^1 \left(\int_0^1 f(x,y,z) \, dz \right)\, dy \right) \, dx$$
where the object on the right may exist but the object on the left may not. However, if $|f|$ is integrable on $[0,1]^3$, then the two must be equal, as one consequence of Fubini's theorem.
Best Answer
The divergence (Gauss-Green) theorem can be used to define the improper integral of the divergence of (weakly) singular vector fields $\mathbf{F}$ with isolated singular points $\mathbf{p}_o=(x_0,y_o,z_0)\in V$. Customarily, the definition goes as follows $$ \begin{split} \int\limits_{V}\nabla\cdot\mathbf{F}(x,y,z)\,\mathrm{d}V& \triangleq \lim_{R\to 0} \Bigg[\,\int\limits_{V\setminus B(\mathbf{p}_o,R)} \nabla\cdot\mathbf{F}(x,y,z)\, \mathrm{d}V - \int\limits_{\partial B(\mathbf{p}_o,R)} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS\Bigg]\\ \\ &\triangleq \int\limits_{\partial V} \mathbf{F}(x,y,z) \cdot \hat{n}\ dS, \end{split}\label{1}\tag{1} $$ where, for the small volume $\delta$, a small ball $B(\mathbf{p}_o,R)$ with radius $R>0$ centered on the singular point of $\mathbf{F}$ is customarily chosen. The definition is clearly consistent if and only if the limits of the two integrals in formula \eqref{1} exist and are finite.
An example.
The most famous example of use of \eqref{1} as a definition is perhaps the calculation of the integral of the divergence of the following field: $$ \begin{split} \mathbf{F}(x,y,z)&=\nabla{\bigg[\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 }\,\bigg]^{-1}}\\ &=\nabla\frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \end{split} $$ where $\mathbf{p}=(x,y,z)\in V$. This vector field is, apart from a multiplicative constant, the gradient of the fundamental solution of the laplacian: therefore, the integral of the divergence of this vector field is zero in every domain $V\subset\Bbb R^3\setminus\mathbf{p}_o$ since $\nabla\cdot\mathbf{F}$ is zero in such domains. However, applying \eqref{1} we have $$ \begin{split} \int\limits_{V}\nabla\cdot\mathbf{F}(\mathbf{p})\,\mathrm{d}V&= -\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \nabla\frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \cdot\hat{n}\, \mathrm{d}S \\ &= -\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \frac{ \partial }{\partial \hat{n}} \frac{1}{|\;\mathbf{p}-\mathbf{p}_o|} \mathrm{d}S\\ &=-\lim_{R\to 0} \int\limits_{ \partial B(\mathbf{p},R)} \frac{ \partial }{\partial r} \frac{1}{r}\mathrm{d}S\\ &=\lim_{R\to 0} \frac{1}{R^2} \int\limits_{ \partial B(\mathbf{p},R)} \mathrm{d}S = 4\pi, \end{split} $$ and thus we can also define the flux of $\mathbf{F}$ throug $\partial V$.
Final notes
[1] A. N. Tikhonov and A. A. Samarskii (1990) [1963], "Equations of mathematical physics", New York: Dover Publications, pp. XVI+765 ISBN 0-486-66422-8, MR0165209, Zbl 0111.29008.