It should be constructed without using 2 stacks. I tried it but I couldn't do it without 2 stacks.
Here's the strategy: we can easily make a PDA that accepts a^n b^n, and we can easily make one that accepts a^n b^2n. Our language is a superset of these languages that also accepts anything with a number of b in between n and 2n. We can make use of nondeterminism to allow this as follows: for every a we put onto the stack, we can nondeterministically decide whether to consume one or two b before popping the a. If our NPDA chooses to consume one each time, we get a^n b^n. If it choose to consume two each time, we get a^n b^2n. If it chooses some of both, we get a number of b in between these extremes. We only accept when we exhaust the input with an empty stack.
Q s S Q' S' Comment
q0 e e qA e Allow empty string to be accepted
q0 a x q0 ax Count a and push onto stack
q0 e x q1 x Transition to counting b
q1 b ax q1 x Mark off a single b for each a
q1 b ax q2 x Mark off the first of two b for this a
q1 e e qA e Allow string in language to be accepted
q2 b x q1 x Mark off the second of two b for this a
In this PDA we have q0 as the initial state and qA as the accepting state. Processing on aabbb:
q0, aabbb, e
-> q0, abbb, a
-> q0, bbb, aa
-> q1, bbb, aa
-> q1, bb, a
-> q2, b, e
-> q1, e, e
-> qA, e, e
Of course there are many parsings that don't lead to qA but in an NPDA we accept if there is at least one.