At the instant of commutation (the voltage reversal), the inductor tries to maintain the exact same current flow. It doesn't try to impose a voltage anywhere, it just forces that current through whatever is connected to it at the time. The voltage that develops is determined by the components connected across the inductor, through which that current must flow.
Take the following as an example. I close a switch to establish a known current through the inductor. I then open the switch to break that current path:
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Initial inductor current (while SW1 is closed) is shown on AM1, and set by R2:
$$ I = \frac{V_1}{R_2} = \frac{12V}{10\Omega} = 1.2A $$
When the switch opens, the inductor will continue to pass 1.2A downwards, through the only path available to it, which is upwards through R1. We can predict what voltage will appear across R1:
$$ V_{AB} = -I \times R_1 = -1.2A \times 100\Omega = -120V $$
This can be seen in a simulation, in which the switch opens after 10μs. This is a plot of voltage \$V_{AB}\$ across the inductor:
Inductors oppose change in current, which is why current remains at 1.2A immediately following the opening of the switch. That can be seen if we plot current:
Resistor R1, therefore, is responsible for whatever voltage will appear between A and B, and inductor L1 is only providing the current, having no influence on voltage at all.
It's the resistor, or diode, or whatever else you connect across the inductor, that determines what voltage will appear there. To obtain a "spike" of 6V across the inductor, choose an appropriate resistance that will develop 6V when 1.2A flows through it:
$$ R_1 = \frac{6V}{1.2A} = 5\Omega $$
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Your question is about the case where a diode is connected across the inductance, not a resistor, so we are interested in finding what voltage will appear in that case instead:
simulate this circuit
We know the current that will flow immediately following switch opening, 1.2A, so we could use the diode equation to reveal the voltage that a diode would develop with 1.2A flowing through it. This is the equation rearranged to make voltage the subject:
$$ V = ηV_T \ln \left(\frac{I}{I_S} + 1\right) $$
Setting parameters roughly according to the 1N4001 diode, thermal voltage \$V_T=25mV\$, saturation current \$I_S=8\times 10^{-11}A\$ and ideality \$\eta=1.5\$, this is what we get:
$$
\begin{aligned}
V &= 1.5 \times 25mV \times \ln \left(\frac{1.2A}{8\times 10^{-11}A} + 1\right) \\ \\
&= 0.88V
\end{aligned}
$$
You already knew that a forward biased diode has somewhere around 0.7V across it, so that should be no surprise. Don't forget that the cathode, node A has the lower potential, since the diode is forward biased, so we expect to see \$V_{AB} = V_A - V_B = -0.88V\$:
That little "wobble" at the end is no doubt due to the diode's junction capacitance resonating with the inductance.
The short answer, then, is that the inductor doesn't choose what voltage it generates, only the current that it attempts to maintain. The voltage is determined by the element through which that current flows, which in the case of a diode is 0.7V or so.
In other words, the inductor does not impose some huge voltage across the diode, it only forces current through it, and the voltage that results will depend on the diode's response to that current, according to its I-V curve.
These same principles apply to your second question. You can use Ohm's law and the diode equation to calculate the total voltage across a series-pair of diode and resistor, to work out what voltage to expect across them both, when the kick occurs.
Combining the two last examples, by putting R1=5Ω and the diode in series, which would then both be momentarily passing 1.2A, we would expect:
$$ V_{AB} = -(0.88V + 6V) = -6.9V $$
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