Lets assume that (in some aritary units of reflectivity) a perfect white has a reflectivity of 100 and perfect black a reflectivity of 0. When you image a narrow white line next to a narrow black line, the image you get won't have a perfect white and a perfet black. It will have perhaps a 90 reflectivity white and a 10 reflectivity black. This is because there is some "smearing" across the lines, so the presence of the white line increases the reflectivity of the black line in the resulting image; while the presence of the black line reduces the reflectivity of the white line in the image. So if the contrast between the white and black lines in the original was 100 (100 - 0) then the contrast between the lines in the image may be only 80 (90 - 10), which is 80% of the original. If you image this a second time through the same imaging system, then you may find that the resulting second-generation image has white lines with reflectivity of 82 and black lines with reflectivity of 18, so the contrast is now only 64, which is 80% of the contrast level of the subject (which was the output of the first imaging process which had contrast of 80). In this case the imaging system has a response of 80% since the contrast of the image is reduced to 80% of its original value by each pass through the imaging system. Re-imaging the output again would result in a contrast of 80% of 64, and so on.
Of course in the case of a negative film, there is an additional inversion that comes into play. I'm ignoring this in the interests of simplicity. Just assume the image I refer to to be the perfect inverse of the film negative that is created.
The extent to which contrast is reduced by a pass through an imaging system is dependant on the spatial frequency (narrowness of the lines) which is measured in lp/mm. The narrower the lines, the greater the reduction in contrast. This is because most contrast reducing effects are most pronounced the closer you are to the edge of the line. Optical effects like diffraction, checmical effects like silver migration (for example due to the silver solvent found in many developers) and imperfections in lenses etc will generally be most noticable close to a high cntrast transition.
The MTF chart shows what percentage of the original contrast is retained in a pass through an imaging system (in this case, from subject to negative film) depending on the spatial frequency of the detail you are trying to image. So when the chart says there is a 70% response at 50 lp/mm it means that when a test chart with lines spaced at 50 lp/mm is imaged, then the resulting image (in this case on negative film) will have 70% of the contrast of the original. In the example you give, if the liones are made narrower so that the spoatial frequency is 150 lp/mm then only 29% of the original contrast is retained.
When a resolution in lp/mm is quoted for an imaging system without specifying the response, then I belivee it is conventional to quote the figure for a 50% response. However marketing being what it is (lies, damn lies and statistics), you should probably not assume this always to be the case.
The question of how much actual resolution you will get depends on what the contrast is on the original subject and what contrast you require in the image. This will tell you what response you require, which can then be converted into a spatial frequency (lp/mm) using the MTF chart.