I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.
1. Why self-inductance is not considered when solving Faraday's law problems
Self inductance should be considered, but is left out for simplicity. So for example, if you have a planar circuit with inductance $L$, resistance $R$, area $A$, and there is a magnetic field of strength $B$ normal to the plane of the circuit, then the EMF is given by $\mathcal{E}=-L \dot{I} - A \dot{B}$.
This means, for example, that if $\dot{B}$ is constant, then, setting $IR=\mathcal{E}$, we find $\dot{I} = -\frac{R}{L} I - \frac{A}{L} \dot{B}$. If the current is $0$ at $t=0$, then for $t>0$ the current is given by $I(t)=-\frac{A}{R} \dot{B} \left(1-\exp(\frac{-t}{L/R}) \right)$. At very late times $t \gg \frac{L}{R}$, the current is $-\frac{A \dot{B}}{R}$, as you would find by ignoring the inductance. However, at early times, the inductance prevents a sudden jump of the current to this value, so there is a factor of $1-\exp(\frac{-t}{L/R})$, which causes a smooth increase in the current.
2. Why an EMF can ever produce a current in a circuit with non-zero self-inductance.
You are worried that EMF caused by the circuit's inductance will prevent any current from flowing. Consider the planar circuit as in part one, and suppose there is a external emf $V$ applied to the circuit (and no longer any external magnetic field). The easiest way to see that current will flow is by making an analogy with classical mechanics: the current $I$ is analogous to a velocty $v$; the resistance is analogous to a drag term, since it represents dissipation; the inductance is like mass, since the inductance opposes a change in the current the same way a mass opposes a change in velocity; and the EMF $V$ is analogous to a force. Now you have no problem believing that if you push on an object in a viscous fluid it will start moving, so you should have no problem believing that a current will start to flow.
To analyze the math, all we have to do is replace $-A \dot{B}$ by $V$ in our previous equations, we find the current is $I(t) = \frac{V}{R} \left(1-\exp(\frac{-t}{L/R}) \right)$, so as before the current increases smoothly from $0$ to its value $\frac{V}{R}$ at $t=\infty$.
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).
Best Answer
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...