I assume what is meant by Faraday's law of induction is what Griffiths refers to as the "universal flux rule", the statement of which can be found in this question. This covers both cases 1) and 2), even though in 1) it is justified by the third Maxwell equation1 and in 2) by the Lorentz force law.
The universal flux rule is a consequence of the third Maxwell equation, the Lorentz force law, and Gauss's law for magnetism (the second Maxwell equation). To the extent that those three laws are fundamental, the universal flux rule is not.
I won't comment on whether the universal flux rule is intuitively true. But the real relationship is given by the derivation of the universal flux rule from the Maxwell equations and the Lorentz force law. You can derive it yourself, but it requires you to either:
- know the form of the Leibniz integral rule for integration over an oriented surface in three dimensions
- be able to derive #1 from the more general statement using differential geometry
- be able to come up with an intuitive sort of argument involving infinitesimal deformations of the loop, like what is shown here.
If you look at the formula for (1), and set $\mathbf{F} = \mathbf{B}$, you see that
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{a} &= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} + \iint_{\Sigma} \mathbf{v}(\nabla \cdot \mathbf{B}) \cdot \mathrm{d}\mathbf{a} - \int_{\partial \Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= - \iint_{\Sigma} \nabla \times \mathbf{E} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= -\int_{\partial \Sigma} \mathbf{E} + \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell
\end{align*}
where we have used the third Maxwell equation, Gauss's law for magnetism, and the Kelvin--Stokes theorem. The final expression on the right hand side is of course the negative emf in the loop, and we recover the universal flux rule.
Observe that the first term, $\iint_\Sigma \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a}$, becomes the electric part of the emf, so if the loop is stationary and the magnetic field changes, then the resulting emf is entirely due to the induced electric field. In contrast, the third term, $-\int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell$, becomes the magnetic part of the emf, so if the magnetic field is constant and the loop moves, then the resulting emf is entirely due to the Lorentz force. In general, when the magnetic field may change and the loop may also move simultaneously, the total emf is the sum of these two contributions.
If you are an undergrad taking a first course in electromagnetism, you should know the statement of the universal flux rule, and you should be able to justify it by working out specific cases using the third Maxwell equation, the Lorentz force law, or some combination thereof, but I can't imagine you would be asked for the proof of the general case from scratch, as given above.
The universal flux rule only applies to the case of an idealized wire, modelled as a continuous one-dimensional closed curve in which current is constrained to flow, that possibly undergoes a continuous deformation. It cannot be used for cases like the Faraday disc. In such cases you will need to go back to the first principles, that is, the third Maxwell equation and the Lorentz force law. There is no shortcut or generalization of the flux rule that you can apply. You should be able to do this on an exam.
1 This equation is also often referred to as "Faraday's law" (which I try to avoid) or the "Maxwell--Faraday equation/law" (which I will also avoid here because of the potential to cause confusion).
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
With a stationary loop of wire and field $\vec B$ varying in time it doesn't seem that you'd be able to exploit the Lorentz force, because it's difficult to introduce the velocity $v$. But another configuration may be more helpful, see picture.
A metallic rod (orange) crosses the magnetic lines (light-blue) with a constant velocity $v$ (blue) in the shown direction. It indeed moves the electrons in the rod as your formula $(1)$ says. The rod glides on two metallic tracks (light-gray) which are connected through another rod (light-gray) to form a closed circuit.
Although the field $B$ is constant, the magnetic flux $\Phi$ varies in time, because
$$\Phi_B = \int_{\scriptstyle_S} \mathbf{B} \cdot \text d\mathbf S \tag{i}$$
In our case is the surface $S$ is rectangular, so we can write the flux as $\Phi_B = B S = B L D$, where $D$ is the length of the rod, and $L$ the length of the metallic tracks on which the rod glides.
For the electromotive force we are interested in the flux derivative with time
$$\mathcal E = -\frac {\text d\Phi_B}{\text d t} = v B D. \tag{ii}$$
because $\text dL/dt = -v$.
Now, I believe that my formula $\text {(ii)}$ and your $(2)$ begin to resemble one another. What misses in my $\text {(ii)}$ is the charges $q$, and what has to appear in your $(2)$ is the rod length $D$, which probably is connected with the element of distance $\text d \vec {\ell}$. Also, since in our case the force $\vec F$ in your $(1)$ is along the rod, and the rod velocity $\vec v$ is perpendicular to $B$, the three vectors $\vec F$, $\vec v$ and $\text d \vec {\ell}$ are mutual perpendicular, s.t. the fact that in my $\text {(ii)}$, the vector product does not appear, is not a concern.
At this point I leave the issue to you. I suggest you to think what is the relationship between work $W$ and potential difference $\mathcal E$, and what is the connection between the integration over the loop in your $(2)$ and the quantity $D$ in my $\text {(ii)}$. Hint: in the grey bars no electromotive force is produced.