How can my private key be revealed if I use the same nonce while generating the signature?
Let me rewrite your question in a different notation, where all lowercase values are integers and uppercase values are points.


The group generator is G (a known constant).
The private key is q, its corresponding public key is Q = qG.
The nonce is n, its corresponding point is R = nG.
The X coordinate of R is r.
The hash function is h(x).
A signature is (r,s), where s is computed as n-1(h(m) + qr).
A signature is valid iff r = x(s-1(h(m)G + rQ)) (where x() stands for "The X coordinate of point ...")


Now for the two signatures it holds that:


s1 = n-1(h(m1) + qr)
s2 = n-1(h(m2) + qr)
s1 - s2 = n-1(h(m1) - h(m2))
n = (s1 - s2)-1(h(m1) - h(m2))


As s1 and s2 are just integers, (s1 - s2)-1 can be trivially computed using a modular inverse; there are no elliptic curve points involved here (over which this problem would be hard).

Once you know n, you can find q by rewriting the first equation:


ns1 = h(m1) + qr
ns1 - h(m1) = qr
q = r-1(ns1 - h(m1))


Substituting this in the earlier equations gives you:


q = (r (s1 - s2))-1(m1s2 - m2s1)