Here is one way to think about it:
When a charged particle travels in a magnetic field, it experiences a force. If the particle is stationary but the field is moving, then in the frame of reference of the field the particle should see the same force.
Now let's take a conductor wound into a coil. In order to increase the magnetic field inside, I could take a dipole magnet and move it close to the coil. As I do so, magnetic field lines cross the conductor, and generate a force on the charge carriers.
It is a convenient trick for figuring out "what goes where" to know that the induced current will flow so as to oppose the magnetic field change that generated it. In the perfect case of a superconductor, this "opposing" is perfect - this is the basis of magnetic levitation. For resistive conductors, the induced current is not quite sufficient to oppose the magnetic field, so some magnetic field is left.
The point is that the flowing of the current is instantaneous - it happens as the magnetic field tries to establish in the coil. So it's not "Apply field in coil. Coil notices, and generates an opposing field. " - instead, it is "Start to apply field in coil. Coil notices and prevents field getting to expected strength".
Not sure if this makes things any clearer...
When you change the flux through a circuit, there are two reasons the flux changes:
1) First, the $\vec{B}$ field in a surface instantaneously spanned by the circuit (at that moment) is changing, in which case there is an electric field in that surface with a circulation $\oint \vec{E}\cdot d\vec{\ell}$ s around the loop that equals $\int -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{a}$, so that:
$$ \oint_{\partial S} \vec{E}\cdot d\vec{\ell}=\int\int_S -\left(\frac{\partial \vec{B}}{\partial t}\right)\cdot d\vec{a}.$$
And this is rightly Faraday's Law (not the "universal" flux rule), because it is the mathematically equivalent integral version of: $$\vec{\nabla}\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}.$$
So Faraday's law says that circulating electric fields cause the $\vec{B}$ field to change (popular sayings get the causality backwards). And get this straight, a circulating electric field is what causes the $\vec{B}$ field to change and the changing $\vec{B}$ field through an instantaneous surface between the circuit is one (of two) things that can make the flux change. The second and third experiments fall in this category. So what is the other reason the flux can change?
2) Second, the circuit itself can have velocity, $\vec{v}$, so the change in location of the circuit in the instantaneous $\vec{B}$ field could result in the $\vec{B}$ field being integrated through a surface whose boundary is changing. In this case (because there are no magnetic monopoles), the change in flux due to the moving circuit equals the circulation $\oint_{\partial S} -\left(\vec{v}\times\vec{B}\right)\cdot d\vec{\ell}.$ The nonmobile charges in the moving circuit are stressed by the magnetic force, but in the quasistatic limit the strain on the nonmobile charges is neglected (and already included in the motion of the circuit) and also in the quasistatic limit the actual motion of the mobile charges differs from the motion of the circuit $\vec{v}$ only by something parallel to the circuit direction $d\vec{\ell}$ so that $\oint_{\partial S} -\left(\vec{v}\times\vec{B}\right)\cdot d\vec{\ell}$ is actually numerically equal to (the negative of) the circulation of the magnetic force per unit charge around the circuit. The first experiment falls in this category. Well, technically the current produces its own $\vec{B}$ field, and it moves so there is a changing $\vec{B}$ field, so there is a little bit of circulating electric field even in the first experiment. This is called self inductance, so the first experiment includes both effects. But it is the only example amongst the three experiments listed that has this second effect where the magnetic force per unit charge is contributing to the emf $\mathscr E$ because the circuit element is moving through a $\vec{B}$ field.
Since these two effects completely determine the change in flux and the change in flux is the sum of these two changes (product rule), the (negative of the) total change in flux is equal to the sum of the circulation of the electric force per unit charge around the circuit and the circulation of the magnetic force per unit charge around the circuit. Their sum is the circulation of the Lorentz Force per unit charge around the circuit, which is the emf, $\mathscr E$, due to electromagnetic forces.
Thus, in quasistatics:
$$\mathscr E=-\frac{d \Phi}{dt}$$
Now, I have to say that I don't see any reason to think the "universal flux rule" actually holds outside quasistatics, since in general charges can move with a velocity other than the velocity of the wire plus a velocity term parallel to the wire. Thus the second effect due to the moving circuit will not always be equal to the circulation of the magnetic force per unit charge around the circuit. But it will be if charges aren't flying off of your circuit and instead the charges only going around it. So you still know when to expect it to hold. In the quasistatic limit, electrostatic forces have time to keep mobile charges flowing through the wire, and electrostatic fields don't contribute to the electromagnetic emf. But it does mean the name ``universal flux rule'' is a misnomer.
Finally as a caveat. I said the electrostatic forces didn't contribute to the emf, but since the circuit is moving, the electric fields responsible for keeping the mobile charges inside the wires (not flying out of the wires) can be non-electrostatic electric fields, which are then what is responsible for the self inductance.
Best Answer
The fact of the matter is that a DC current induces a constant magnetic field, except for the instant at which it commences and ceases. If a square wave, for example, were applied to the primary of a transformer, the secondary would produce a series of spikes of alternating polarity (as in an automotive ignition coil). Voltage can be transformed by a DC-to-DC converter, which works by generating an AC current, applying it to a transformer, and rectifying the output of the secondary, but the cost is higher than when using a transformer. (This is an oversimplification, but the details are not the question here.) Transformers and induction motors require an AC current. Take the case of a high voltage (up to 1,000,000 volts) long distance transmission line. It operates at the high voltage to reduce losses, but the voltage is obviously unsuitable for either generation or local distribution, thus requiring transformers to change the voltage. It is much more economically feasible to use an AC system from one end to the other. (Nowadays, since the advent of cost effective and reliable high-power semiconductors, some high voltage transmission lines are operated at DC, requiring conversion at both ends.) So, the answer to your question about why AC is used: it's cheaper.
For an interesting take on the matter, look up the battle between Edison Electric Light, which advocated DC distribution and Westinghouse Electric Company, which chose AC.