If you have two valid signatures using the same private key where k' = 2k then:

From each message we can derive the z value (hash of the message) so:

First message and signature (m, r, s, z)
Second message and signature (m', r', s', z')

Therefore:  ks = z + rd_A and k's' = z' + r'd_A

Therefore:  (sk - z)/r = (s'k' - z')/r'

But in this case k' = 2k so:

(sk - z)/r = (2s'k - z')/r'

So all you have to do is solve for k.  All the other values:  s, z, r, s', z', and r' are all known.

rr'(sk - z)/r = rr'(2s'k - z')/r'

r'(sk - z) = r(2s'k - z')

r'sk - r'z = 2rs'k - rz'

r'sk - r'z +rz' = 2rs'k

k = (r'sk - r'z +rz')/2rs'

Once you know k you can simply calculate the private key, d_A = (sk - z)/r

I still do not see what your two signatures have to do with the BTC at 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx

These two things:  how to solve for the private key when you know k' = 2k and the BTC stored at 1FvUkW8thcqG6HP7gAvAjcR52fR7CYodBx seem to be unrelated, right?