Decrypting Ciphertext with partial Key Fragment using LFSR and Berlekamp-Massey
TL;DR:
If you have a similar question, then use the online Berlakamp-Massey algorithm to get the minimal polynomial by providing the given keystream, then get the C++ code from github and then replace your tap positions, seed, and ciphertext and then compile and run.

The answer is attraction
How to get it
What is given: We have given the following information;

Key fragment
seed
ciphertext

Attack idea: The key fragment is given to calculate the shortest LFSR that outputs the given key fragment. Once you get the taps, initialize the LFSR with the seed then decrypt the ciphertext stream.
Steps in details;

Use Berlakamp-Massay algorithm to find a shortest LFSR (need not to be unique). An online version is used.

The output is given as $x^{16} + x^{12} + x^3 + x^1 + 1$

As noted on the online site the size of the LFSR is 16 and the tap positions are
0,1,3,12
  xxxxxxxxxxxxxxxx &lt;-|
  || |       |       |
  ++-+-------+-------|  
  01 3       12


Initialize the LSFR of size 16 with the seed and this tap positions. Run the LFSR and get the keystream;
11001010011101010101100010111110110000000011011000100011110101100001101101111101

X-or the keystream with the ciphertext

Convert the plaintext to ASCII to get
attraction



note 1: if you x-or the 16-bit seed with the ciphertext you will get the at. This has a very low probability to contain two letters from the English Alphabet. If we assume that only small English letters are used then the probability is $26^2/256^2 \approx 1.031\%$. This was the actual clue int the question. And, normally, real stream ciphers don't start immediately form the initialization. They step for a certain amount from the beginning.
note 2: A sample C++ code can be found in the github.