I found it quite hard to follow both the question and the answer, so I am going to state a few relevant things and paraphrase the answers of @LDC3 to add some clarity.
Chemical shift (ppm) and peak area (integral)
These are independent quantities. One tells you about the environment (shift) the protons are in, and the other tells you how many protons are in that environment, compared to the number of protons in all the other environments.
That the areas need to be considered as a ratio instead of absolute values is important: if the molecule was symmetric, with an ethyl group either side of a ketone e.g. C₅H₁₀O you would see a peak ratio of 3:2 (two environments), but there would actually be 6 protons in the first environment and 4 in the second. This is a bit like empirical vs molecular formulae.
Some tips for deducing structures
1) Start with 'characteristic' shifts. These are usually the lowest and highest shifts.
Here the 4.1 ppm shift is characteristic of ester protons, i.e. -(C=O)OC H₂R (R can also be a proton too!) where, much like the data sheet above, bold is to tell you which protons are being referred to in a more complicated structure. The shift at 0.9 ppm is characteristic of protons which are attached to a standard saturated alkyl chain. This tells you that those protons aren't very near any oxygen, nitrogen or other electron withdrawing or electron donating groups.
2) Try and explain the other shifts using functional groups that you have identified from your characteristic shifts
The shift at 2.0 ppm is likely to be alkyl chain protons which are next to an electron withdrawing group, which conveniently you have already identified as an ester group by looking at your characteristic shifts! In this case, that means you can put the 0.9 ppm shifted protons at one end of your structure, because there is nowhere else to put them.
3) Check your peak assignment makes sense when you look at your peak areas
The last thing to check is which end of the molecule the 0.9 ppm shifted protons go on. i.e. is the molecule methyl propanoate or ethyl ethanoate? To do this you can look at the peak areas. For this example it is quite easy to check because there are 3 protons at 4.1 ppm, so you know that you have a methyl ester group -(C=O)OC H₃. Our R group from earlier is an H after all! The resulting structure is the one shown in the example: methyl propanoate.