I'm not sure what a total derivative has to do with this. Why don't we see how a change of variables is going to affect a double integral?
Suppose there is a rectangle ##R## in the arbitrary ##uv##-plane. Suppose further the lower left corner of the rectangle is at the point ##(u_0, v_0)##, and the dimensions of the rectangle are ##\Delta u## for the width and ##\Delta v## for the height respectively.
Now let's define an invertible transformation ##T: (u, v) \rightarrow (x, y)## such that the rectangle ##R## in the ##uv##-plane can be mapped to a region ##R'## in the ##xy##-plane. This transformation from the ##uv##-plane to the ##xy##-plane will be given by some ##x = g(u, v)## and ##y = h(u, v)##. For example, the lower left corner of ##R## can be mapped to the boundary of ##R'## by using ##T## like so:
$$T(u_0, v_0) = (x_0, y_0)$$
$$x_0 = g(u_0, v_0), \space y_0 = h(u_0, v_0)$$
Now, define a vector function ##\vec r(u, v)## to be the position vector of the image of the point ##(u, v)##:
$$\vec r(u, v) = g(u, v) \hat i + h(u, v) \hat j$$
Note the equation for the bottom side of the rectangle ##R## in the ##uv##-plane is given by ##v = v_0##, and the equation for the left side of the rectangle ##R## in the ##uv##-plane is given by ##u = u_0##. The image of the bottom side of ##R## in the ##xy##-plane is given by ##\vec r(u , v_0)##, and the image of the left side of ##R## in the ##xy##-plane is given by ##\vec r(u_0 , v)##.
The tangent vector at ##(x_0, y_0)## to the image ##\vec r(u , v_0)## is given by:
$$\vec r_u(u, v) = g_u(u_0, v_0) \hat i + h_u(u_0, v_0) \hat j = x_u \hat i + y_u \hat j$$
Similarly, the tangent vector at ##(x_0, y_0)## to the image ##\vec r(u_0 , v)## is given by:
$$\vec r_v(u, v) = g_v(u_0, v_0) \hat i + h_v(u_0, v_0) \hat j = x_v \hat i + y_v \hat j$$
We can approximate the region ##R'## in the ##xy##-plane by a parallelogram determined by the secant vectors:
$$\vec a = \vec r(u_0 + \Delta u, v_0) - \vec r(u_0, v_0) ≈ \Delta u \vec r_u$$
$$\vec b = \vec r(u_0, v_0 + \Delta v) - \vec r(u_0, v_0) ≈ \Delta v \vec r_v$$
So to determine the area of ##R'##, we must determine the area of the parallelogram formed by the secant vectors. So we compute:
$$\Delta A_{R'} ≈ \left| (\Delta u \vec r_u) \times (\Delta v \vec r_v) \right| = \left| \vec r_u \times \vec r_v \right| (\Delta u \Delta v)$$
Where we have pulled out ##\Delta u \Delta v## because it is constant. Computing the magnitude of the cross product we obtain:
$$\left| \vec r_u \times \vec r_v \right| = x_u y_v - x_v y_u$$
So we may write:
$$\Delta A_{R'} ≈ [x_u y_v - x_v y_u] (\Delta u \Delta v)$$
Where ##x_u y_v - x_v y_u## can be determined by evaluating ##g_u(u_0, v_0), h_v(u_0, v_0), g_v(u_0, v_0)##, and ##h_u(u_0, v_0)## respectively.
Now that we have formalized all of that, divide the region ##R## in the ##uv##-plane into infinitesimally small rectangles ##R_{ij}##. The images of the ##R_{ij}## in the ##xy##-plane are represented by the rectangles ##R_{ij}'##. Applying the approximation ##\Delta A_{R'}## to each ##R_{ij}'##, we can approximate the double integral of a function ##f## over ##R'## like so:
$$\iint_{R'} f(x, y) dA ≈ \sum_{i = 1}^m \sum_{j = 1}^n f(x_i, y_j) \Delta A_{R'} ≈ \sum_{i = 1}^m \sum_{j = 1}^n f(g(u_i, v_j), h(u_i, v_j)) [x_u y_v - x_v y_u] (\Delta u \Delta v)$$
Notice this looks like a typical Riemann sum. Now as ##m \to \infty## and ##n \to \infty##, the double sum converges to a double integral over ##R##:
$$\displaystyle \lim_{m \to \infty} \displaystyle \lim_{n \to \infty} \sum_{i = 1}^m \sum_{j = 1}^n f(g(u_i, v_j), h(u_i, v_j)) [x_u y_v - x_v y_u] (\Delta u \Delta v) = \iint_R f(g(u, v), h(u, v)) [x_u y_v - x_v y_u] \space dudv$$
Where we usually write the Jacobian ##J = [x_u y_v - x_v y_u]##. So we can finally conclude:
$$\iint_{R'} f(x,y) \space dA = \iint_{R} f(g(u, v), h(u, v)) \space J \space dudv$$
This argument for an arbitrary ##(u, v)## space applies to polar ##(r, \theta)## space as well. In fact, this argument will apply for any kind of other invertible transformation ##T##.