On the application of chaos in signal processing
A typical communications system is composed of the following parts:

Message (Information Source) --&gt; Encoding --&gt; Modulation --&gt; Channel --&gt; Demodulation --&gt; Decoding --&gt; Recovered Message


A "chaotic modulator" is just another way of doing the Modulation building block. This is depicted in Figure 1 of the paper that is cited in the question.

The input to the Modulator is a binary sequence and the output of it is a digital (or directly analog, depending on the system) signal.

Therefore, to clarify (rather than answer) the first issue: As per Figure 1, the bs are essentially your Message (or Message after it has been encoded for whatever purpose (e.g. error correction).

A typical Modulator's job is to map binary symbols to carrier symbols with the carrier signal being one that is suitable for transmission effectively through the Channel. In a typical system, we can literally see this mapping in the Constellation Diagram.

In the case of a chaotic modulator though, binary symbols are mapped to different states of the chaotic modulator. This is mentioned in the introduction of the paper as:


  Rather than using structured signals, such as rectangular  pulses or
  sinusoids, to denote ‘0’s and ‘l’s, these communications systems embed
  the information in the time evolution, or dynamics, of the transmitted
  signal.


A simpler way to think about this is to take some difference equation (such as f) and produce a few values with some initial condition x0 and a few values for some initial condition x1. If the two time series are distinct enough, then perhaps it is possible to recover which of the two x0,x1 might have generated this time series. In this case, of going from time series to some initial condition, we are looking at some sort of Demodulator.

Now, chaotic of not, a difference equation provides a model. A model imposes a set of constraints and some sort of structure to the signal. This is what is exploited here (please keep in mind that we are talking about a pseudo-chaotic system). So, what if I take look at the difference equation and realise that I can express it as a typical digital filter?

Again, to clarify, rather than answer, this is what Drake &amp; Williams did with "Pseudo-chaos for direct sequence spread spectrum communications". From the paper linked in the question:


  More  importantly, Drake  and Williams  [7] have  shown  that the
  sawtooth  map's  output is equivalent  to the response of a linear
  filter to a binary sequence.


Having clarified this, it is also important to highlight Figure 2 (and 3 to some extent) from the original paper which shows that by inversing the filtering operation it is effectively possible to do Modulation and Demodulation.

This, hopefully, clarifies the "representation" missunderstanding.

Finally, I would like to stress the pseudo-chaotic nature of the generated signal which comes as a result of the modifications that are also explained in section 2, from "In practice this is not a possible because this filter..." (sic) onwards. 

I would also like to recommend that you take a brief look at Pseudo-random number generators, the Linear congruential generator and finally the linear feedback shift register (LSFR) for examples of seemingly linear "representations" that still produce rather complex sequences and then taking a brief look at the Dyadic Mapping where it is also mentioned that:


  The name bit shift map arises because, if the value of an iterate is
  written in binary notation, the next iterate is obtained by shifting
  the binary point one bit to the right, and if the bit to the left of
  the new binary point is a "one", replacing it with a zero.


Hope this helps.

EDIT:

In view of additional comment:


  If the chaotic map is denoted by the state variable $x$, then these
  $b$ are the bits. Further, are the $b$′s obtained from the binary
  expansion of some numeric value of the iterate of chaotic map $x$?


Here is an alternative way to go about this: 

Let's say that this is not a modulator. Ultimately, $x$ is the output from the mapping. We start with some initial value $x_0$ and produce $x_1$ which we send back into the map and get back $x_2$ and again through the map to get $x_3$ and back through it once more to get a $x_4$ and so on to get $x_n$. Since we are operating on a computer with numbers of finite wordlength, $x$ contains DIGITAL values (in the decimal system) with a well defined minimum and maximum. So, if $x$ is a byte, then the min value is 0 and the maximum value is 255.

Each one of these $x_n$ decimal values has a binary representation as well. Therefore:

$x_0$ is equivalent to some binary number $0100101$

$x_1$ is equivalent to some binary number $1001010$

$x_2$ is equivalent to some binary number $0010100$

And eventually...

$x_n$ will be equivalent to some $b_m b_{m+1} b_{m+2} ...$ where some $b \in {0,1}$

Therefore, it is possible to have a binary string of digits (0010001101...) for a pseudo-chaotic sequence $x$.

Let us now ask the reverse question: What if all we have is the binary sequence. Can we go from the binary sequence to the output it would produce?

To do this, you would have to establish a relationship between the $b$ (binary sequence) and $x$ (decimal sequence / digital signal). It turns out that this is possible via the work of reference 7 as mentioned above, through $h$.

To recap and put it in the context of communications:

$b$ is your Message. If the message was "ABC", its ASC representation would be 65,66,67 and the binary representation of that would be "010000010100001001000011". This is your $b$.

$x$ is the output from the Modulator with pseudo-chaotic dynamics.

$h$ is now the tool by which the map is applied to the data.

(Please see figures 2,3 in the original paper).


  How can the convolution of $b$ with impulse give rise to a chaotic
  map, $x$?


Again...The input to the modulator is a binary sequence ($b$). Each $b$ is either $0$ or $1$. The output from the modulator is a digital signal ($x$) (decimal values of some wordlength). 

If we take a look at:

$x[n] = \sum_{k=-\infty}^{\infty}{b[n-k] \times h[k]}$

We realise that:


$x$ is a weighted sum of $b$. 
To produce one $x_n$, we need to sum many bits from $b$. How many? $k$ many.
Obviously, $k$ cannot remain infinity in a practical system. That's where truncation of $h$ comes in.


I hope this helps.