From the perspective of the lines of flux, if the coil is at 90 degrees to these lines, the coils area is maximized i.e. the maximum number of lines of flux flow thru the coil and the induced emf would be maximum. If the plane of the coil were rotated 90 degrees to be in line with the lines of flux, the effective area of the coil (from the perspective of the lines of flux) is zero and there will be no induced emf.
At any point in between, the "effective area" changes as a sin(angle) function, where "angle" is 90 degrees when the coil is totally across the field lines and zero when it is in line.
Does this help?
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.
Best Answer
The apparent contradiction arises because you're confusing \$ H\$ with \$ B\$. Remember that \$ H\$ is insensitive to whether or not a magnetic material is present (analogously to how \$ D\$ is insensitive to whether a dielectric is present for electric fields). So the \$ H\$ field between \$ dh\$ is not stronger than the \$ H\$ field between \$ bf\$ (which is not completely zero, and is only neglected because the magnetic flux in the core is much stronger, and the flux is usually what we care about in practical scenarios because it features in Faraday's law). The \$ B\$ field is of course much stronger between \$ dh\$ but as I said, Ampere's law concerns \$ H\$, not \$ B\$. If you want to apply Ampere's law with \$ B\$ rather than \$ H\$, then you need to take into account the permeabilities of the core and the air as you do your line integrals and this difference will account for the apparent contradiction.