Why does K+ going out of the cell cause hyperpolarization?

Adding K+ on the outside of membrane is not the same as transfering K+ from inside to outside. When K+ is added to outside(hyperkalemia) the RMP becomes less negative(depolarisation) When K+ is transfered from in to out RMP becomes more negative(re/hyperpolarisation) We should not forget that most of the intracellular anions are nonpermeable protiens, so when K+ is transfered to outside they pull harder, and when K+ added outside, the pull decreases

I agree this is confusing.

My understanding is that the Nernst potential is given by the equilibrium concentrations of the given ion, i.e. the concentrations when the cell is unstimulated. Therefore, even during an action potential, the Nernst potentials for sodium and potassium definitionally don't change.

So I would remove the "Nernst potential" part from your definitions.

Now as for how I make sense of the action potential graph:

A positive transmembrane voltage indicates an outwards force on positive ions. Therefore, as sodium enters the cell, the transmembrane voltage increases because the inside of the cell is more positively charged.

Similarly, a negative transmembrane voltage indicates an inwards force on positive ions. So as positively-charged potassium leaves the cell, the transmembrane voltage decreases because the interior is now more negatively charged.

To corroborate, here's a definition from Raz 2013 that includes Nernst potential: "Depolarization is a positive change from the resting potential achieved by increased permeability to an ion with a Nernst potential above the RBP. Ion flux drives the membrane potential closer to the Nernst potential of this ion." Note how this indicates that Nernst potential doesn't change.

Also, while I think your definitions make more sense (i.e. they're more literal), this quote implies to me that "depolarization" has shifted from its historical definition to now just mean "an increase in transmembrane voltage". Alas, I'm also somewhat confused about this topic, but hopefully this answer is nonetheless correct and helpful.

Here is how I think of the issue. First, keep in mind over the course of the action potential, ion concentrations on both the outside and inside of the neuron remain relatively unchanged. You can think of the Nernst potential as a charged battery, and they keep their concentrations relatively constant. Currents will flow, and the voltage will change, but this effects very few ions at a time, and does not effect the bulk concentration (See section 2.6 here). This is because any small change in concentration near the membrane (where voltage is measured) will quickly equalize with the surrounding bulk solution via diffusion.

Second keep in mind that the Nernst potential is an electro-chemical potential. Thus for potassium in particular, the chemical potential will overpower the electric potential driving potassium out of the cell, making the driving voltage of potassium negative.

So, as you state, the Nernst potential of sodium is $60$ $mV$ and for potassium is $-90$ $mV$. In your example there is a rest potential of $-70$ $mV$ this is given by the GHK potential mentioned in the comments. That means at rest, the balance between the sodium and potassium potentials is $-70$ $mV$. Here is the equation for reference. Note that this equation reduces to the Nernst potential when we consider only one ion.

$$ V = frac{RT}{F} ln(frac{g_{Na}[Na_{out}]+g_{K}[K_{out}]}{g_{Na}[Na_{in}]+g_{K}[K_{in}]})$$

When the action potential starts, the depolarization is driven by the opening of sodium channels, that is $g_{Na}$ gets larger. Thus, when these channels open sodium rushes out and the cell voltage attempts to become more like the sodium Nernst potential, not zero. You can see that in the equation for the GHK voltage equation as the sodium conductance dominates the equation.

Near the peak of the action potential, the sodium channels close, and the potassium channels open. That is $g_{Na}$ gets small, and $g_{K}$ gets large. This means that now the potassium conductance dominates the equation, and the neuron's voltage becomes more like the $-90$ $mV$ Nernst potential. Then the repolarization occurs when the potassium channels close, and $g_{Na}$ and $g_K$ are back to there intial closed values, bring the rest potential back to $-70$ $mV$.

**note I am oversimplifying how the gates change in time, but I think this will help clear up your confusions.