why does ${a^nb^n}$ fit the pumping lemma for context-free languages?
This language satisfies the conditions of the Pumping lemma. (By the way, your question title is wrong: you are not asking why it is context-free; you are asking why it satisfies the conditions of the lemma, which it does.)

Take the string s=a^mb^m and assume m&gt;0. Don't use p or n, otherwise it is confusing, because those letters are used in the lemma. Now let v="a", y="b", x="", u=a^(m-1), z=b^(m-1). 

You can definitely pump it as much as you want. In other words, you take a string such as aaaaaabbbbb, then you pick the two middle letters: aaaa ab bbbb. Then you pump them: aaaa aaabbb bbbb. Still in the language.