I'll use the notation in the references, thus the question's privkey1 (resp. privkey2, nonce1, nonce2) are noted $x_1$ (resp. $x_2$, $k_1$, $k_2$). The curve order is noted $p$ (rather than the usual $n$).
The question mention "(…) 4 transactions for 2 pubkeys with the same r1 and r2", but gives different values for r1 and r2. I'll read instead: 4 ECDSA signatures matching 2 private/public keys pairs using only 2 nonces, in the following arrangement:
| hash | nonce | privkey | signature | equation$\pmod p$ |
|---|---|---|---|---|
| $h_1$ | $k_1$ | $x_1$ | $(r_1,s_1)$ | $s_1k_1≡r_1x_1+h_1$ |
| $h_2$ | $k_1$ | $x_2$ | $(r_1,s_2)$ | $s_2k_1≡r_1x_2+h_2$ |
| $h_3$ | $k_2$ | $x_1$ | $(r_2,s_3)$ | $s_3k_2≡r_2x_1+h_3$ |
| $h_4$ | $k_2$ | $x_2$ | $(r_2,s_4)$ | $s_4k_2≡r_2x_2+h_4$ |
The numbers given for $x_i$, $k_i$, $r_i$, $h_j$, $s_j$ for $i\in\{1,2\}$ and $j\in\{1,2,3,4\}$ all are in $(1,p)$; verify the equations; and $r_i$ is the function of $k_i$ prescribed by ECDSA on secp256k1, that is $r_i$ is the X coordinate of $k_i\,G$ reduced modulo $p$ (the reduction seldom makes a difference, and this is no exception).
But contrary to the references, it holds $s_1s_4≡s_2s_3\pmod p$. That prevents applying the method in the references to find $x_1$ and $x_2$, which requires that $r_1r_2(s_1s_4-s_2s_3)$ be invertible modulo $p$.
Under the assumption $h_1$, $h_2$, $h_3$, $h_4$ (or the corresponding signed messages) and $x_1$, $x_2$, $k_1$, $k_2$ are arbitrary, with the signatures derived from that, there is no reason $s_1s_4≡s_2s_3\pmod p$ would hold. That it holds makes the system of 4 equations with 4 unknowns $x_1$, $x_2$, $k_1$, $k_2$ impossible to solve from the signatures and hashes alone; we'd need some additional relation involving at least one of the unknowns $x_1$, $x_2$, $k_1$, $k_2$, but none is stated.
One possibility is that there's none to be found. The whole thing could be a decoy, a joke, perhaps a scam (which that comment suggests). One way to build the question's numbers would be that $s_4$ is computed as $s_1^{-1}\,s_2\,s_3\bmod p$, then $h_4$ is computed from $s_4$ rather than as the hash of some message.