About random self-reducibility of DDH
You are absolutely right! The random self-reducibility goes in the other direction, and this variant of the DDH assumption does not follow from it. I have no idea what the author was thinking when he wrote it :-). In any case, the paper has now been updated in ePrint and fixed. Thank you for catching this.

I include the proof of this variant here:

Consider the following DDH variant that states that for every probabilistic-polynomial time non-uniform distinguisher $D$ there exists a negligible function $\mu$ such that
\begin{equation}
\left|\Pr[D(\mathbb{G},q,g_0,g_1,(g_0)^r,(g_1)^r)=1] - \Pr[D(\mathbb{G},q,g_0,g_1,(g_0)^r,(g_1)^{r+1})=1]\right| \leq \mu(n)
\label{eq:ddh}
\end{equation}
where $\mathbb{G}$ is a group of prime order $q$ with generator $g_0$, $g_1\in\mathbb{G}$ is a random group element, and $r\in\mathbb{Z}_q$ is randomly chosen. This assumption can be proven to be to be true if the standard DDH assumption holds. In order to see this, observe that by the standard DDH assumption
\begin{equation}
\left|\Pr[D(\mathbb{G},q,g_0,g_1,(g_0)^r,(g_1)^r)=1] - \Pr[D(\mathbb{G},q,g_0,g_1,(g_0)^r,(g_1)^s)=1]\right| \leq \mu(n)
\label{eq:ddh1}
\end{equation}
where $g_1\in\mathbb{G}$ and $r,s\in\mathbb{Z}_q$ are randomly chosen. In addition, observe that the distribution over $(g_0,g_1,(g_0)^r,(g_1)^s)$ is identical to the distribution over $(g_0,g_1,(g_0)^r,(g_1)^{s+1})$. Thus, a straightforward reduction to the standard DDH assumption gives that 
\begin{equation}
\left|\Pr[D(\mathbb{G},q,g_0,g_1,(g_0)^r,(g_1)^s)=1] - \Pr[D(\mathbb{G},q,g_0,g_1,(g_0)^r,(g_1)^{r+1})=1]\right| \leq \mu(n).
\label{eq:ddh2}
\end{equation}
In order to see this, a distinguisher $D'$ receiving $(g_0,g_1,h_0,h_1)$ can run $D$ on $(g_0,g_1,h_0,h_1\cdot g_1)$. If $D'$ received $(g_0,g_1,(g_0)^r,(g_1)^r)$ then it generates a tuple of the form $(g_0,g_1,(g_0)^r,(g_1)^{r+1})$, and if $D'$ received $(g_0,g_1,(g_0)^r,(g_1)^s)$ then it generates a tuple of the form $(g_0,g_1,(g_0)^r,(g_1)^{s+1})$, which as we have mentioned is identical to $(g_0,g_1,(g_0)^r,(g_1)^s)$. Thus, if $D$ can distinguish with non-negligible probability in this last equation, then $D'$ can use $D$ to solve the standard DDH problem. Combining these last two equations, we obtain that the first equation (which is the DDH variant assumption) holds.