Assuming I'm reading the post correctly, the question seems to be "why/how is the standard notion of relative entropy related to the given expression for the max-relative entropy?
Consider the standard definition of (classical) relative entropy between two probability distribution $P,Q$:
$$D(P\|Q)=\sum_i P_i \log\left(\frac{P_i}{Q_i}\right).$$
You can read this as the expectation value of the quantity $\log(P_i/Q_i)$, over the distribution $P$.
Suppose instead of taking the average you take the maximum value of this quantity, defining
$$D_{\rm max}(P\|Q) = \max_i \log\left(\frac{P_i}{Q_i}\right).$$
There is another equivalent way to write this: note that $\max_i c_i$ is equal to the smallest $\lambda$ such that $c_i\le \lambda$ for all $i$. In other words, you could rewrite $D_{\rm max}$ as
$$D_{\rm max}(P\|Q) = \inf\{\lambda\in\mathbb{R}:\,\,\log(P_i/Q_i)\le\lambda,\,\,\forall i\}
\\ = \inf\{\lambda\in\mathbb{R}: \,\, P\le 2^\lambda Q\}
= \log\inf\{\eta\ge0:\,\,P\le \eta Q\}. $$
Note that in the last two expressions we're writing $P\le 2^\lambda Q$, where $P,Q$ are vectors here, to indicate $P_i\le 2^\lambda Q_i$ for all $i$.
In the quantum case you can write the relative entropy as
$$D(\rho\|\sigma)=\operatorname{tr}(\rho\log\rho)-\operatorname{tr}(\rho\log\sigma) =
\sum_{i,j} p_i |\langle p_i|q_j\rangle|^2 \log\left(\frac{p_i}{q_j}\right),$$
as also discussed here.
A little trickier than the classical expression, but it's still an expectation value of $\log(p_i/q_j)$, though now wrt to a joint probability distribution depending on both $\rho,\sigma$, which doesn't reduce to the classical case unless $[\rho,\sigma]=0$.
Now, I'm actually not sure whether there is a way to go from this to the expression for $D_{\rm max}$ as cleanly as the above argument for classical states. However, observe that
$\inf \{\lambda\in\mathbb{R}:\,\,\rho\le 2^\lambda \sigma\}$ is the smallest real $\lambda$ such that $\rho\le 2^\lambda \sigma$, which means the smallest $\lambda$ such that $2^\lambda\sigma-\rho\ge0$. An equivalent condition is that, for all positive semidefinite operators $Z\ge0$, we have
$$\langle Z,\rho\rangle\le 2^\lambda \langle Z,\sigma\rangle,$$
where we're using the trace inner product: $\langle Z,A\rangle\equiv \operatorname{tr}(Z^\dagger A)$. Therefore
$$\inf \{\lambda\in\mathbb{R}:\,\,\rho\le 2^\lambda \sigma\} =
\log\inf \{\eta\ge0:\,\,\frac{\langle Z,\rho\rangle}{\langle Z,\sigma\rangle}\le \eta, \,\,\forall Z\ge0\} \\
= \log\sup_{Z\ge0} \frac{\langle Z,\rho\rangle}{\langle Z,\sigma\rangle}
= \sup_{Z\ge0}\log \frac{\langle Z,\rho\rangle}{\langle Z,\sigma\rangle}. $$
While we didn't get to this directly from converting averages to max in the expression for $D(\rho\|\sigma)$, this is clearly quite similar to what we have for $D_{\rm max}(P\|Q)$ classically, except that we're replacing the max over outcomes to its quantum counterpart: a max over the log-ratio of the possible probabilities measuring the two states in a common basis.
See the second lecture of Watrous' QIT notes for more information.
