Leakage in power spectral density estimation
Could somebody explain what does this mean? Why does a window leak
  power into adjacent frequency bins?


You can think of frequency bins as imposing an artificial frequency grid which allows only certain frequencies to fit nicely within it (producing a sharp peak) and all other frequencies are smeared across adjacent bins thus producing spectral leakage.

This means that only those frequencies that are centered exactly on bin frequency will not produce leakage.

Since the bandwidth of a bin (i.e frequency resolution) is simply the 
sampling rate divided by the dft length, we can conclude that only frequencies that are exact integer multiples of the frequency resolution will not produce spectral leakage.

For example, if the sampling frequency is 8000 Hz and dft length is 1024, the frequency resolution is computed as 

 frequency_resolution = 8000 / 1024 = 7,8125 Hz


So, if your signal contains frequencies which are integer multiples of 7,8125, you'll get
no leakage (e.g 1*7,8125, 2*7,8125, 3*7,8125 and so on)

If, however, your signal contains a frequency that is somewhere between 2 adjacent bins, you'll get spectral leakage, which I think is intuitive because no single bin can
fit that frequency component exactly.

UPDATE (trying to answer the comments below): 

It's true that windowing does introduce some additional leakage on its own. But even if you don't use windowing at all (which corresponds to a rectangular window), you'll still get spectral leakage (provided that the frequency under examination is not an integer multiple of the frequency resolution). 
In fact, since using a rectangular window (no window at all) produces the worst leakage, using overlapping windows other than the rectangular helps reduce the spectral leakage.

So, why do we have spectral leakage in the first place?

Remember that DFT assumes that a signal is periodic. Since the frequency resolution is  fixed (e.g 7,8125 Hz in the example above) for every bin, which in this case corresponds to exactly 1024 samples, we can see that a single bin can hold exactly one cycle of a 7,8125 Hz frequency , 2 cycles of 15,625 Hz frequency, 3 cycles of 24,4375Hz frequency and so on. This is true for every bin, which means that there'll be no discontinuities either within a bin or between adjacent bins. 

However, if we now change our frequency to something that fits between two bins, things change. This new frequency will take either more (or fewer) than 1024 samples (one or more cycles), which means that we'll get discontinuities either within a bin or between neighbouring bins and this in turn creates spectral leakage. 

Hope this helps.