I don't think Hilmar's answer is very good as it interprets the DFT within a specific application context. That confuses issues.
The DFT is a tranform that works on a set of N samples. The samples are presumed to be evenly spaced in their domain on a finite interval of samples is called a frame. It may be time, it may be distance, or it could even be another dimension. The bin values tell you how close your samples are to corresponding to that bin. The important thing to remember is each bin corresponds to a sinusoidal which has a frequency of the bin index measured in units of cycles per frame. What happens outside the frame is out of scope for the DFT. Saying "The DFT assumes the interval is infinitely repeating" is misleading without any constructive benefit. What is true is that an inverse DFT can be extended, and it will form a repeating pattern.
The concept of "energy" is only valid is a subset of applications. "Leakage" is not about energy at all, it is about representation of points in multidimensional coordinate systems. But that discussion is a bit mathematical.
What it seems you are stuggling with having to do with definitions on particular applications and what is the significance of the number of sample points on your frequency values and the possible resolutions. Those concepts are straightforward with a understanding of the DFT devoid of application details.
Suppose you have a given signal over a fixed interval. If that signal has three cycles in the frame, the parameters of the fundamental will be found in bin 3 (zero based indexing), and its harmonics will be in bin 6, 9, 12, etc. The number of sample points you use will determine the number of bins there are. The more bins there are the higher up your halfway point (called Nyquist) is. If the frequency of your signal (or one of its harmonics) exceeds the Nyquist value, then it will "wrap around" and look like a different frequency. This is called an alias.
For instance, suppose you had N=16. Then the Nyquist bin is 8. A signal will 10 cycles per frame will land in bin 10, but its mirror image value lands in bin 6, so if you are looking at the DFT you would say, "Hey there is a 6 cycles per frame signal in there", when in fact it is 10. Increasing your sample count would fix that. For N=32, the 10 bin would still have a value, but now the mirror image is in 22. That's why you have to sample at least twice the rate of the highest frequency you want to find to have it land in the lower half.
BTW, every DFT is exact, aka lossless.
The sampling rate is what ties the units of the DFT to the units of your application. N has units of samples per frame. $f_s$ is the conventional symbol for samples per application unit, usually seconds, so the unit is samples per second, designated with Hz. This is a little misleading as the sampling is better thought of as a rate instead of a frequency. Anyway, using T for your interval length, measured in seconds, you have.
$$ f_s \left(\frac{samples}{second}\right) = \frac{ N \left(\frac{samples}{frame}\right) }{ T \left(\frac{seconds}{frame}\right) } $$
Of course, you can switch that around, say:
$$ N = f_s \cdot T $$
To interpret the bin values, $k$ is usually used for the bin index, so the frequency associated with a bin is $k$ cycles per frame. Hence:
$$ \frac{ k \left(\frac{cycles}{frame}\right) }{ T \left(\frac{seconds}{frame}\right) } = k/T \left(\frac{cycles}{second}\right) $$
Therefore the "bin spacing", i.e. the difference in frequency between two adjacent bins, is $1/T$, but the units are still cycles per second.
Now, from above, $ T = N / f_s $, so the bin spacing can also be called $ f_s / N $.
And just a bit more insight...
Your Nyquist frequency is always going to be 2 samples per cycle for any N.
$$ \frac{ 2\pi \left(\frac{radians}{cycle}\right) }{ 2 \left(\frac{samples}{cycle}\right) } = \pi \left(\frac{radians}{sample}\right) $$
It is also helpful to picture the DFT bins arranged around the unit circle in the complex plane. The DC bin on the 1, and the Nyquist is on the -1. The Nyquist is the same distance from the DC whether you go clockwise or counter-clockwise. The bins are indexed from 0 to N/2 across the top for even N, and 0 to (N-1)/2 for odd N. There is only a Nyquist bin for even N. The bottom half is better indexed with negative values($-k$), but usually comes back as the upper part of the results($N-k$). The definition doesn't care and they are aliases.
The mirror image, aka conjugate symmetry, is only true for real valued signals. Ultimately it works the way it does because:
$$ \cos( \theta ) = \frac{ e^{i\theta} + e^{-i\theta} }{2} $$
But that's a longer story.