KVL is a circuital law. You need to define a circuit path, let's call it $\Gamma$, that joins your circuital components. If you can find a path $\Gamma$ that does *not* link the changing magnetic flux, then you can use KVL on that path and the sum of all voltages across your components' terminals will end up being zero.
#### Ordinary KVL
When there is no magnetic flux present in your circuit and components we can express 'ordinary KVL' as:
$$
\oint_{\Gamma}\vec{E}.\vec{dl}=0
$$
In fact, we can decompose this closed path into the partial paths going through each component; after accounting for all the signs, we will end up with EMF when we go through a battery, an ohmic voltage drop when we go through a resistor and a capacitor voltage when we go through a capacitor...
$$
\int_{battery}\vec{E}.\vec{dl} + \int_{resistor}\vec{E}.\vec{dl} + \int_{capacitor}\vec{E}.\vec{dl} =0
$$
[![Figure: circuit with battery, R and C][1]][1]
...and their sum (with appropriate signs) will be zero, as stated by '**ordinary KVL**'. As a side note, in this conservative setting, all integrals relative to resistors and capacitors are path-independent and we can either integrate along the path going through the components or jumping across the terminals.
#### Extended KVL
Whenever a changing magnetic flux is present and is linked by the circuit path $\Gamma$, 'ordinary KVL' no longer applies because the circulation of E is no longer zero, as stated by Faraday's law:
$$
\oint_{\Gamma=\partial\Sigma}\vec{E}.\vec{dl}=-\frac{d}{dt}\iint_\Sigma \vec{B}.\vec{dS}=-\frac{d\phi}{dt}
$$
[![Figure: circuit with path linking changing flux][2]][2]
In *most* (but not all) cases, though, it is possible to define a new circuit path $\Gamma'$ in such a way that no appreciable (changing) magnetic flux is linked by it; in order to do that we will have to introduce one or more new components - that I will call 'magnetic components' for short - that will conceal the magnetic flux inside them. We end up with a new circuit that will link all non-magnetic components as before (by either going through them or jumping at their terminals, as you prefer) but will definitely skip the interior of the magnetic ones by 'jumping' at their terminals.
[![Figure: circuit with path skipping the magnetic component][3]][3]
Since the new circuit path $\Gamma'$ joins all components without linking any of the changing flux we can write:
$$
\int_{\Gamma'}\vec{E}.\vec{dl}=0
$$
This is an expression of 'extended KVL' and, as before, we can break up the integral into partial segments that represent the various components, but will always 'skip' the magnetic components by jumping at their terminals in order to leave the magnetic flux region out of the circuit's path.
$$
\int_{battery}\vec{E}.\vec{dl} + \int_{resistor}\vec{E}.\vec{dl} + \int_{capacitor}\vec{E}.\vec{dl}+ \int_{jump_{A->B}}\vec{E}.\vec{dl} =0
$$
Now, here comes the neat trick: we express the path integral along the 'jump' at the magnetic component's terminals in terms of the inner - and hidden from the circuit's path $\Gamma'$ - changing magnetic flux. I am going to do it in the simplest of magnetic component: a coil linking an externally generated variable magnetic flux (not a self-inductor, and neither the secondary of a transformer, but a magnetic sensor).
Figure: the single coil with external flux - note directions in the paths
How do we compute the path integral of E.dl along the jump at its terminals? Simple: we compute the circulation of E.dl along the closed path formed by said jump and the coil's conductor invoking Faraday's law
$$
\int_{coil_{A->B}}\vec{E}.\vec{dl}+\int_{jump_{B->A}}\vec{E}.\vec{dl}=\oint_{\partial\Sigma}\vec{E}.\vec{dl}=-\frac{d\phi}{dt}
$$
and then we subtract the path integral of E.dl along the coil's conductor: if the conductor is perfect, the electric field inside of it will be zero and the path integral will be... zero. By reversing the endpoints of the path (because we want $\Gamma'$ to proceed from A to B, like $\Gamma$ did, we end up with
$$
\int_{jump_{A->B}}\vec{E}.\vec{dl}=-\int_{jump_{B->A}}\vec{E}.\vec{dl}=+\frac{d\phi}{dt}
$$
And we can pretend this is a voltage drop just like the ones for resistors and capacitors.
Let's confront the two ways to treat a circuit with lumped magnetic components: The purist will apply Faraday's law directly
Figure 5+3 = 8
The practitioner will fold back to the 'extended KVL' for the amended path $\Gamma'$ in the following form
Figure 5+3-8=0
#### No KVL
Now let's go back to the derivation of the voltage *across* our coil. Observe carefully: if you accept the definition of voltage as the path integral of the *total* electric field along a given path (see [my answer here][4] for the definition of voltage in line with the IEC definitions) you have one value of voltage **across** the coil's terminals (let's call it VL), and a different value of voltage **along** the coil itself. This difference makes sense because the electric field is no longer conservative and so its path integral are in general path-dependent.
KVL is no longer applicable **inside a magnetic component**. But it still works when applied to the modified circuit path $\Gamma'$ . (This is why Feynman, in chapter 22 of the second volume of his lectures, insists that in order to use circuital simplifications no magnetic fields must escape the lumped components.)
Are there **circuits that are not 'KVL-amendable'**? The answer is yes: one example of such a circuit is a ring with two opposing resistors (or capacitors, or inductors, or a mix of the above) that are *required* to be on the opposite sides of a changing magnetic flux region.
Figure: Lewin's ring
This constraint on the position of the resistors with respect to the flux region forces the circuit path $\Gamma$ to link the changing flux and makes it impossible to find a new circuit path $\Gamma'$ that excludes the magnetic flux region. This is the circuit used by Walter Lewin in his much discussed superdemo, and it is an example of unlumpable circuit for which KVL does not apply. And I want to stress: the V in K**V**L is for voltage and, according to the IEC definition, voltage is the path integral of the total electric field (and not its conservative component $\vec{E}_c=\vec{E}-(-\partial\vec{A}/\partial{t})$ alone, which is responsible for the scalar potential difference.)
**References:**
Ramo, Whinnery, VanDuzer, "Fields and Waves in Communication Electronics", Wiley. Chapter 4 "The electromagnetics of circuits".
Purcell, Morin, "Electricity and Magnetism", Cambridge
Feynman, "Feynman's Lectures on Physics", Addison Wesley. Volume 2 , Chapter 22.
[1]: https://i.sstatic.net/TDYwP.png
[2]: https://i.sstatic.net/bFNmp.png
[3]: https://i.sstatic.net/aPyNI.png
[4]: https://physics.stackexchange.com/questions/667777/is-voltage-and-electric-potential-actually-the-same-thing-if-not-why/733761#733761