Let $$ T(u, v, w)=x(u, v, w) \hat{i} +y(u, v, w)\hat{j} + z(u, v, w)\hat{k}$$ be a continuously differentiable bijection.
Then $$ \int \int \int_D f(x, y, z) dx dy dz=\int \int \int _D f \circ T(u, v, w) \det{J} du dv dw, $$ where $$ J=\left[\begin{matrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w} \end{matrix}\right]. $$
The main goal is to show that $$ \Delta V= \Delta x \Delta y \Delta z = \det{J} \Delta u \Delta v \Delta w .$$
Attempt of proof:
Using local approximation we get $$\Delta T \approx J \left[\begin{matrix} \Delta u\\ \Delta v\\ \Delta w \end{matrix}\right] \\ \left[\begin{matrix} \Delta x\left(u{,}\ v{,}\ w\right)\\ \Delta y\left(u{,}\ v{,}\ w\right)\\ \Delta z\left(u{,}\ v{,}\ w\right) \end{matrix}\right]\approx\left[\begin{matrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w} \end{matrix}\right]\left[\begin{matrix} \Delta u\\ \Delta v\\ \Delta w \end{matrix}\right] \\ \left[\begin{matrix} \Delta x\left(u{,}\ v{,}\ w\right)\\ \Delta y\left(u{,}\ v{,}\ w\right)\\ \Delta z\left(u{,}\ v{,}\ w\right) \end{matrix}\right]\approx\left[\begin{matrix} \frac{\partial x}{\partial u}\Delta u+\frac{\partial x}{\partial v}\Delta v+\frac{\partial x}{\partial w}\Delta w\\ \frac{\partial y}{\partial u}\Delta u+\frac{\partial y}{\partial v}\Delta v+\frac{\partial y}{\partial w}\Delta w\\ \frac{\partial z}{\partial u}\Delta u+\frac{\partial z}{\partial v}\Delta v+\frac{\partial z}{\partial w}\Delta w \end{matrix}\right]$$
So now if we read row by row $$ \Delta x= \dots, \Delta y= \dots, \ \mathrm{and \ so \ on}, $$ and calculate $$ \Delta V= \Delta x \Delta y \Delta z,$$ we end up in a different result. So why doesn't it work using this approach?
I think your attempt is fine, the only thing that you need to formalize is what "$\approx$" means, and what it means is that you choose on Taylor series up to first derivative.
Here is a simpler way to not make all the computations. It should be easier (and equivalent) to see $T$ one component at a time, that is, verify this three cases: $$T(u,v,w)=x(u,v,w)\hat{i}+u\hat{j}+w\hat{k}$$ $$T(u,v,w)=u\hat{i}+y(u,v,w)\hat{j}+w\hat{k}$$ $$T(u,v,w)=u\hat{i}+v\hat{j}+z(u,v,w)\hat{k}$$ And by symmetry, it is enough to show only one, choose the first one.
In this case $$J=\left[\begin{matrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w}\end{matrix}\right]= \left[\begin{matrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\ 0&1&0\\ 0&0&1\end{matrix}\right]$$ Hence, $$\Delta x =\frac{\partial x}{\partial u}\Delta u +\frac{\partial x}{\partial u} \Delta v +\frac{\partial x}{\partial u} \Delta w, \quad \Delta y=\Delta v,\quad \Delta z=\Delta w $$
Finally, $$\Delta V=\Delta x \Delta y \Delta z\approx \left(\frac{\partial x}{\partial u}\Delta u +\frac{\partial x}{\partial u} \Delta v +\frac{\partial x}{\partial u} \Delta w\right)\Delta v \Delta w \approx \frac{\partial x}{\partial u}\Delta u\Delta v \Delta w =\operatorname{det}J \Delta u \Delta v \Delta w$$
Where terms of second order ($\Delta v^2$ and $\Delta w^2$) vanish in our approximation.