What kind of attack does the current brokenness of SHA-1 allow?
That's true. There should not be any publicly known attack on SHA-1 that allows for a given $d1$ to find a $d2$ such that $h(d1) = h(d2)$.

SHA-1 is vulnerable  for finding a pair of $d1$ and $d2$ such that $h(d1) = h(d2)$.

The same, however, applies to MD5, which is unusable for SSL/TLS certificates and a successful attack has been performed. So, this kind of collision can allow (under some circumstances) more that you might think.

The notable difference between SHA-1 and MD5 attacks is the cost of such attacks. For MD5, the attack is very fast. For SHA-1, the attack would be very expensive, but for some groups (e.g. governments) feasible.

(There might be also some more differences that I am not aware of. However, both of them use the Merkle–Damgård construction, which implies some common security characteristics.)