Faraday's law of induction clarifications - Area change over time
What confuses me is that the flux derivative is inside the surface
  integral.


The partial derivative inside the integral comes from Leibniz Integral Rule (detailed below).

Consider generalized form of Maxwell-Faraday Equation:

$$\oint E \cdot d\ell = -\int_\Sigma \frac{\partial B}{\partial t} \cdot dA$$

This is true for any path \$\partial \Sigma\$, which is any closed-contour bounds the surface \$\Sigma\$.

Now remember generalized form of Leibniz Integral Rule:

$$\frac{d}{dt}\int^{a(x)}_{b(x)} f(x, t) \ dt = f(x, b(x)) \ \frac{d}{dt}b(x) - f(x, a(x)) \ 
 \frac{d}{dt} a(x) + \int^{a(x)}_{b(x)} \frac{\partial}{\partial t}f(x, t) \ dt$$

Did you notice that the bounds of the integral above are not constants?

The right-side term of the generalized Maxwell-Faraday Equation is a surface integral (and, of course, integral around \$\partial \Sigma\$ is a line integral) and the partial derivative inside this integral indicates that any \$\partial \Sigma\$ path is time-dependent. That's why we write Maxwell-Faraday Equation in generalized form because we cannot guarantee that any \$\partial \Sigma\$ is constant.

Now let's look at Leibniz Rule again when the bounds are constants. The first two term of the right-side becomes zero and the integral takes its own special form:

$$\frac{d}{dt}\int^{a}_{b} f(x, t) \ dt = \int^{a}_{b} \frac{\partial}{\partial t}f(x, t) \ dt$$

From this, we can make a conclusion: If the path \$\partial \Sigma\$, which bounds the surface \$\Sigma\$, does not change over time, Maxwell-Faraday Equation turns into:

$$\oint E \cdot d\ell = -\frac{d}{dt}\int_\Sigma B \cdot dA$$

NOTE: I wrote Leibniz Integral Rule for single dimension and made explanations over it just to make things simpler, but the same thing applies for higher dimensions.