Count down from "Infinity"

Jelly, 15 10 bytes

-5 bytes thanks to @Dennis (convert directly from base 256 after ordinal cast)

Oḅ⁹µBCḄµÐĿ

TryItOnline!

How?

Oḅ⁹µBCḄµÐĿ - Main link: s                     e.g. "Inf"
O          - cast to ordinals                 e.g. [73,110,102]
 ḅ⁹        - convert from base 256 to integer e.g. 4812390
   µ   µ   - monadic chain separations
    B      -     convert to binary
     C     -     complement
      Ḅ    -     convert to integer
        ÐĿ - loop until no longer unique and collect results 

Python 2, 89 82 77 76 75 bytes

n=0
for c in input():n=n<<8|ord(c)
while 1:print n;n^=2**n.bit_length()-n/n

Test it on Ideone.

How it works

After initializing n to 0, the second line performs the string-to-integer conversion specified in the challenges as follows.

In each step, n is shifted 8 units to the left, then bitwise OR-ed with the the code point of the next character c. For input Inf, this goes as follows.

n                                  0
a = n<<8                           0
b = 'I'                      1001001
n = a ^ b                    1001001
a = n<<8             100100100000000
b = 'n'                      1101110
n = a ^ b            100100101101110
a = n<<8     10010010110111000000000
b = 'f'                      1100110
n = a ^ b    10010010110111001100110

Now we're ready to generate the output. To invert the bits of n, we proceed as follows.

First, we compute the bits in n's binary representation without leading zeroes. Let's call the result k. Then, we compute the k^k power of 2, which has k+1 binary digits: a single 1, followed by k 0's. We subtract 1 from the result, yielding a number composed of k ones, which we then XOR with n to invert its bits. For input inf this goes as follows.

n         4812390   10010010110111001100110
k              23 
t = 2**k           100000000000000000000000
t -= 1              11111111111111111111111
n ^= t    3576217    1101101001000110011001
k              22
t = 2**k            10000000000000000000000
t -= 1               1111111111111111111111
n ^= t     618086      10010110111001100110
.
.
.
n               6                       110
k               3
t = 2**k                               1000
t -= 1                                  111
n ^= t          1                         1
k               1
t = 2**k                                 10
t -= 1                                    1
n ^= t          0                         0

On additional hurdle in the implementation is that we have to print n before the first step, after the last step, and in all steps in between. Python doesn't have do-while loops and a single print statement costs 8 bytes, so we do the following instead.

In the straightforward implementation of the update step, i.e.,

while n:print n;n^=2**n.bit_length()-1
print n

we replace the loop with an infinite one (while 1) and compute the 1 in the loop as n/n. This is equivalent while n > 0.

Once n = 0, we stay in the loop, print the state once more, then try to update it. However, 0/0 triggers a ZeroDivisionError, breaking out of the loop and exiting with an error. Note that this causes stray output to STDERR, which is allowed by default.

JavaScript, 82 bytes

Saved a byte thanks to @Arnuald

for(y of prompt(n=0))n=n<<8|y.charCodeAt()
for(;alert(n)|n;)for(i=1;i<=n;i*=2)n^=i

One of the very few times when a full program outperforms a function (and ES6 does not outperform ES5)...

The above supports up to 4-letter words. Add 4 bytes to support up to 6-letter words:

for(y of prompt(n=0))n=n*256+y.charCodeAt()
for(;alert(n)|n;n=i-n-1)for(i=1;i<=n;)i*=2