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.
It is, as I think you have decided, the net flux that is responsible for the net induced emf. However, we can write the net emf as
$$\mathscr E = -\frac{d\Phi}{dt}=-\frac{d(\Phi_{ext}+\Phi_{ind})}{dt} = -\frac{d\Phi_{ext}}{dt}+\left(-\frac{d\Phi_{ind}}{dt} \right)= \mathscr E_{ext}+\mathscr E_{ind}$$
Here, $\Phi_{ext}$ is the flux from outside the circuit and $\mathscr E_{ext}$ the emf that can be attributed to the rate of change of that flux. $\Phi_{ind}$ is the flux arising from the induced current at some instant and $\mathscr E_{ind}$ is the emf that can be attributed to the rate of change of that flux.
In many textbook treatments, $\mathscr E_{ind}$ is not at first mentioned. It is often dealt with later under the heading of self induction, an effect that comes into play only when the circuit is complete.
This two-stage treatment of the induced emf may be worrying for the alert student!
Best Answer
The induced current in your triangular conductor would be the same as in a circle of the same area if $\Phi$ changes in this area. Thats what the law says, since $\Phi=B\cdot A$, so $\frac{d \phi}{dt}=A\cdot\frac{dB}{dt}$