What is the significance of 8 bits in the context of an ADC0808?
The short answer is it represents the resolution of the analog to digital converter (ADC).

An ADC solves the problem of getting a digital device (something that only deals in 0's and 1's) to understand an arbitrary analog voltage. The 8 bits represent a binary number between 0 and 255, which are 256 unique numbers. The chip uses a conversion method called "successive approximation" to translate an analog voltage into a binary value between 0 and 255.

Let's say your ADC's reference voltage is 5V. The ADC will split up its 256 possible values evenly across 0V to 5V. More specifically, an analog voltage of 0V will read as 0 (00000000 in binary) and 5V will read as 255 (11111111 in binary). A voltage between 0V and 5V will read as a linear function between 0 and 255. For example, half of the reference voltage, 2.5V, will read as half of 255, or 128 (10000000 in binary). 

The smallest possible voltage change an 8-bit ADC can measure is called its resolution. The resolution can be calculated by dividing the reference voltage by total number of unique binary values. So for our example above:
$$5V/256 = 0.0195V$$
This means that each increment of the lowest significant bit equates to an increase of 19.5mV. You can work out what analog voltage each 8-bit value corresponds to:
0000 0000 = 0.0V
0000 0001 = 0.0195V
0000 0010 = 0.0390V
0000 0011 = 0.0585V
....
1111 1100 = 4.9415V
1111 1101 = 4.9610V
1111 1110 = 4.9805V
1111 1111 = 5.0V  

You can see from these calculations that the more bits, the higher the resolution. If the ADC was, say, 12-bit instead of 8-bit, your resolution would be:
$$5V/4096 = 0.00122V$$
So you could measure much smaller changes in voltage and therefore get a much more accurate measurement. Of course, with higher resolution comes higher cost and longer sampling time.