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.
In electrostatics, the electric field in a wire loop is conservative (why?) so an electron which makes a full turn around wire loses as much energy as it gains. In other words the emf of the circuit is zero, if I understand correctly
Electrostatic field is conservative because if we move a charge from point A to B, the work done is irrespective of the path taken.
The easiest way to prove this is by taking a point $q$ charge, taking it away to some point and then bringing it back to initial point (without increasing its kinetic energy).
For electrostatic field:
$W_{ext}=-q\Delta V=-q(\oint_{closed-loop}\vec E \cdot {d\vec{l})}=-q(-\frac{\mathrm{d}\phi }{\mathrm{d} t})=0 \\ \Rightarrow W=0$
This implies that irrespective of the path taken, work done is always $0$ in an electrostatic field when charge is brought back to initial position since $\frac{d \phi}{dt} $ is always $0$.
For induced electric field:
$W_{ext}=-q\Delta V=-q(\oint_{closed-loop}\vec E \cdot {d\vec{l}})=-q(-\frac{\mathrm{d}\phi }{\mathrm{d} t}) \neq 0 \\ \Rightarrow W \neq 0$
Here $\frac{d\phi}{dt}$ depend on the path taken.
Hence electrostatic field is conservative, but induced electric field is not.
In otherwords the emf of the circuit is zero, if I understand correctly.
Since there is no battery, there is no EMF. In the above case, we are making the charge move around. So there is no need of EMF.
If you were asking about a circuit with a battery, then the EMF is provided by the battery.
However, apparently in a circuit with an induced emf, an electron which makes one full loop can gain energy. So as electrons continue to travel around the circuit, they gain more and more energy. In otherwords, the induced emf is due to an induced electric field in the wire which is nonconservative. Does this make any sense?
The work done in changing the magnetic field is transferred to the electrons in the conductor, which gives them energy to move around. This is a consequence of Law of Conservation of Energy.
Now, main question:
... However, doesn't this contradict our definition (*) of induced emf?**
No. Because it is not rigorous to use Faraday's Law here in the first place.
$\oint_{c}\vec E \cdot {d\vec{l}}=-\frac{d}{dt}\int_{s} \vec B \cdot {d\vec{l}} $
is the common version of Faradays's Law. It is rigorously correct only if $\vec E$ represents the electric field in the rest frame of each segment $d\vec l$ of the path of integration. This is definitely not true is the case of motional EMF.
It is necessary to note that only a time-varying magnetic field induces a circulating non-conservative electric field in the rest frame of the laboratory. A changing flux does not induce an electric field. Hence in case of motional EMF, no electric field is induced in the laboratory frame(where rod is in motion). Hence technically $\oint_{c}\vec E \cdot {d\vec{l}} = 0$ here.
A potential difference is maintained across the rod by the EMF and Lorentz force maintains that EMF. Using $-\frac{d}{dt}\int_{s} \vec B \cdot {d\vec{l}} $ here is just a convenient way of calculating the EMF.
Source: A Student's Guide to Maxwell's Equations -- Daniel Fleisch
Best Answer
This is where it's a good time to "converse with the math". Let's look at the equation:
$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$
which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:
$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$
Thus, what your question is, is essentially, asking "how can you have negative work". Looking at the above equation and recalling how a dot product behaves, there's only one way: $\mathbf{F}$ and the element of displacement $d\mathbf{r}$ must be aimed at cross purposes with each other. In the case of the integral for $\mathcal{E}_\mathrm{ind}$, the same goes only with $\mathbf{E}$ and $d\mathbf{l}$.
Hence, the answer to your question is: when $\mathbf{E}$ points opposite to $d\mathbf{l}$.
But what does that mean? Well, the key here is that we have to think a little more closely about the work formula. I believe what you are imagining it means is "the work done by the force as the force pulls the particle along with it". It actually is more general - in a work integral, the particle can be moved in any direction, including against the force. Of course, to make that motion happen in real life, you need to supply a source of contrary force, but that doesn't change the maths. This is why, say, in a more elementary example, you can talk of "negative work" done by the gravitational force when you lift an object off the floor.
(Why is it defined that way? Well, for one, because we often can't solve for the "real" trajectory the particle follows! If we stipulated that as a precondition, it would make work an extremely non-trivial concept!)
Where your mistake lies in, then, is in missing that. The displacement around the wire $d\mathbf{l}$ that we use to describe emf is not the displacement that necessarily occurs in reality to a real positive charge (after all, in many applications we aren't "really" dealing with positive charges anyways!).
Rather, it is a hypothetical one where we imagine that we grab a charge and move it around all the way through the circuit in a specific, fixed direction, and ask what the work - whether positive or negative - done by the electric force for that movement is. If we ask about the work done in the actual movements of charges, we will get a different answer, and yes, this one will always be positive, at least provided we don't get into looking at the situation in too-fine detail.
From an intuitive point of view, if you want to figure when the force is doing "positive" work and when it's doing "negative" work, imagine that you can feel the electric force tugging on the positive charge in your hand as you move it through the circuit. When you feel it helping you, i.e. the tug is with the motion of your hand, at that moment (i.e. that small increment $d\mathbf{l}$) the electric force is doing positive work, and you are doing negative work (to retard the charge, if you were to try and not naturally speed up your hand as you'd likely tend to, of course). When you feel it is fighting you, i.e. the tug is against the motion of your hand, the electric force is doing negative work, and you are doing positive work (to help it against the contrary pull). Total negative work, and hence negative emf, will be if, in moving the charge, you had to fight more often than flow.