Provable security of cryptographic hash functions
First this, a nit about the hash function: one can trivially create a collision by appending a 0 value; an appended 0 does not change the hash value, and hence that produces a collision.

If we say that collisions are required to have at least one place there two colliding messages differ with at least one having a nonzero value, then we can show it.  Here is how, given a black box which finds a collision, can perform discrete logs (with good probability).

Suppose we have the elements $G$ and $H$, and which to find the value $k = log_G H$, that is, the solution to $kG = H$.

What we do is select $h_i = r_i G + s_i H$, where $r_i, s_i$ are random values.  Then, present these $h_i$ values (along with the rest of the parameters) to the black box.

The black box will produce a collision, that is, two messages $m, m'$ with $\sum m_i h_i = \sum m'_i h_i$.  Rewriting this, we get $(\sum (m_i - m'_i)r_i)G + (\sum (m_i - m'_i)s_i) H = 0$, or $aG + bH = 0$, for values $a, b$ which we can compute (as we know the $m_i, m'_i, r_i, s_i$ values)

As the $r_i, s_i$ parameters are random values unknown to the blackbox (and at least one of the $m_i - m'_i$ values will be nonzero), it is unlikely that $b$ will be 0; if it is not, then we have $k = -ab^{-1}$