From Raul Bragado, head greenkeeper at La Manga Resort:

When you work out the extractions of the different grass species and use different nitrogen doses in different climates then the grams per square metre of potasium or phosphorus is worked out with the ratio of the foliar analysis? Meaning 4/2/1? So if you feed the grass with 25 g/m of N the need of K is F= K mlsn + extractions K (1/2 N) - K soil? F= 5.2 + 12. 5 - 11 = 6.7 for a soil with 11 g/m?

That's correct. Here's how this works, and why. 

We want to ensure the concentration of an element in the soil remains at or above the MLSN guideline. Let's call that guideline level x. By doing a soil test, we can measure the concentration of an element in the soil. Let's call the soil test level z.  The grass is going to use that element over time, and that use is a function of how much the grass grows, with the growth controlled by temperature and by the amount of nitrogen supplied. Let's call the amount the grass uses y. 

The total amount of the element required is the guideline amount (x) plus the amount the grass will use (y). The amount of that element that we have is the soil test level (z). To find the amount required as fertilizer, we need to look at the difference between the amount required and the amount actually present. To express this in an equation, as Raul did, we can use F to denote the fertilizer required and write the equation as:

\[F = x + y - z\]

In cool-season grasses, the nitrogen to phosphorus to potassium ratio in the leaves is about 8:1:4, because the leaves will usually have about 4% N, 0.5% P, and 2% K. So if we apply 25 g N/m², we can expect that the K use will be half of that, at 12.5 g K/m². y in this case would be 12.5.

The amount of K we want to have in the soil is 37 ppm or more. This reflects the new (preview here) MLSN guideline for K. Assuming we have a sand rootzone, with a bulk density of 1.5 g/cm³, and that we are working with a 10 cm rootzone depth, then 37 ppm in the 3-dimensional space of the rootzone is equivalent to 5.5 g/m² in the 2-dimensional space of the surface for uptake or fertilizer application purposes. We need to express x, y, and z in the same units for the equation to give the correct result for F, so let's express x as 5.5.

In the soil test, let's say we have K at 74 ppm. To convert that to g/m², we use the same conversion factor of 1 g/m² = 6.7 ppm, and that gives 11 g/m² for z. Now we can find the fertilizer requirement F. 

F = 5.5 + 12.5 - 11 = 7 g/m²

In Raul's example, he calculated 6.7, and I got an answer of 7. The difference is due to the update in the MLSN guidelines.

Manilagrass (Zoysia matrella) in Kyushu will use less than 15 g N/m2/year

 I really like this approach because it accounts for many practical variables of greenkeeping. For example, if we have a different grass species, such as seashore paspalum, the N:K ratio in the leaves will be about 1:1. So in that case, if we applied 18 g N/m², we could expect the grass to require 18 g K/m². In that way, this approach adjusts to the grass species.

Also, if we had seashore paspalum in Bangkok, the N use might be 32 g N/m²/year. In Okinawa, where it is much cooler, the N use might be 18 g N/m²/year. The amount of K and P and other elements used will adjust accordingly based on location. 

Manilagrass (Zoysia matrella) in central Thailand will use about 30 g N/m2/year

Also, the MLSN guidelines are updated utilizing results from the Global Soil Survey, to ensure that these guideline levels are self-correcting and are based on the nutrient levels found in soils producing good-performing turf all around the world.