Generate m-sequences

Haskell, 88 82 bytes

r@(h:t)%m=h:(t++[mod(sum[v|(v,1)<-zip r m])2])%m
l#x=take(2^l-1)$(1:(0<$[2..l]))%x

I'm using input method 1 (but in reverse order, because I'm shifting to the left) together with L as an explicit parameter.

Usage example (test case #2): 4 # [1,1,0,0] -> [1,0,0,0,1,0,0,1,1,0,1,0,1,1,1].

How it works: % does the work. First parameter is the content of the shift register, second parameter is the feedback configuration. # starts the whole process by building the initial register content, calling % and taking the first 2^L-1 elements.

main function #:

                1:0<$[2..l]      -- initial register content: one 1 followed
                                 -- by L-1 0s
                            %x   -- call the helper function with the
                                 -- feedback configuration
take(2^l-1)                      -- and take the first 2^L-1 elements

helper function % (bind r to the whole register set, h to the first register a0
and t to register a1..aL. Bind m to the feedback config): 

h:                               -- next element is h, followed by
                         %m      -- a recursive call of %, where the new register is
   t++                           -- t followed by
           [v|(v,1)<-zip r m]    -- the elements of l, where the corresponding
                                 -- elements of m are "1"
      mod(sum ...          )2    -- xored    

Pyth, 19 bytes

eM.u+xF*VQNPN+m0tQ1

Try it online: Demonstration or Test Suite

Explanation:

eM.u+xF*VQNPN+m0tQ1   implicit: Q = input list in format 1
              m0tQ    create a list with len(Q)-1 zeros
             +    1   and append a one at the end
                      this is the starting sequence
  .u                  apply the following statement repeatedly to the starting seq. N:
       *VQN              vectorized multiplication of Q and N
     xF                  xor all this values
    +      PN            remove the last value of N and prepend the new calculated one
                      .u stops, when it detects a cycle and returns a list of
                      all intermediate states of N
eM                    take the last element of each state and print this list

Jelly, 18 bytes

æḟ2µ&³B^/+ḤµḤ¡BḣḤṖ

This takes a single integer, encoding a₀ in its MSB and a_L-1 in its LSB.

The test cases become 1, 12, 18 and 48 in this format. Try it online!

How it works

æḟ2µ&³B^/+ḤµḤ¡BḣḤṖ  Main link. Argument: n (binary-encoded taps)

æḟ2                 Round n down to k = 2 ** (L - 1), the nearest power of 2.
   µ                Begin a new, monadic link. Argument: r = k
    &³              Bitwise AND r with n.
      B             Convert to binary.
       ^/           Bitwise XOR all binary digits.
          Ḥ         Unhalve; yield 2r.
         +          Add the results to left and right.
           µ        Convert the previous chain into a single link.
            Ḥ       Unhalve; yield 2k = 2 ** L.
             ¡      Execute the link 2k times, updating r in each iteration.
              B     Convert the final value of r to binary.
               ḣḤ   Keep only the first 2k binary digits.
                 Ṗ  Pop; discard the last digit.