The supremum $q$ of the quantity $q(G)$ you are interested in, over the class of all finite graphs, is at least $\frac13$.

For the time being, I do neither *know* whether $q$ is *larger* than $\frac13$, nor whether the value $\frac13$ can be *attained* by any finite graph.

Here are some details.

It can be proved that there is a sequence of finite graphs on which the quantity converges to $\frac13$.

This sequence consists of *triangle-free*, three-colorable, Cayley graphs only: the sequence $\mathrm{And}_t$ of *Andrásfai graphs* (cf. e.g. the book of Godsil and Royle on algebraic graph theory).

Let $q_{\mathrm{absolute}}(G)$ denote the graph invariant (FiniteGraphs)$\longrightarrow$ $\mathbb{N}$ you defined.

Let $q(G) := q_{\mathrm{relative}}(G) := \frac{1}{\lvert G\rvert} q_{\mathrm{absolute}}(G)$ the quantity about which you asked how large it can get when $G$ ranges over all finite graphs.

It can be proved that the supremum of $q(G)$ over the class of all graphs is at least $\frac13$.

Since $\mathrm{And}_t$ is triangle-free, i.e., $\omega(\mathrm{And}_t)=2$, a *transversal of the maximum cliques* is equivalent to a cover of the edges by vertices (usually, and somewhat counterintuitively, called a *vertex cover* in contemporary graph theory texts).

So for any triangle-free graph $G$, the quantity $\min_T\lvert T\rvert$, in your sense, *without* the penalty-subtrahend, is just $\tau(G)$, the *covering number* of $G$.

This will now be used to give a rough lower bound on your quantity $q(G)$.
The penality-subtrahend will just be *estimated away*, making use of the fact that Andrásfai graphs have relatively *small independence number*, using a bound in terms of the independence number (I decided *not to think* about how much the bound of $\frac13$ can be improved if one does *not* do this; this would require an analysis of the *structure* of the set of all independent sets of $\mathrm{And}_t$, which should be a straightforward task).

For every natural number $t$, the $t$-th Andrásfai graph $\mathrm{And}_t$ has

$\lvert \mathrm{And}_t\rvert = 3t-1$,

$\alpha(\mathrm{And}_t) = t = \tfrac13(\lvert \mathrm{And}_t\rvert+1)$,

$\tau(\mathrm{And}_t) = 2t-1 = \lvert \mathrm{And}_t\rvert - t$.

We can now argue as follows, abbreviating $n_t:=\lvert\mathrm{And}_t\rvert$,

$q$ $=$ $\sup_{\text{allfinitegraphs}} q(G)$

$\geq$

$\sup_{t\in\mathbb{Z}_{\geq 2}}\frac{1}{n_t}(\min_T \min_{A\subseteq T} \lvert T\rvert - \lvert A\rvert)(\mathrm{And}_t)$

$\geq$

$\sup_{t\in\mathbb{Z}_{\geq 2}}\frac{1}{n_t}(- \alpha( \mathrm{And}_t ) + (\min_T \lvert T\rvert )(\mathrm{And}_t) )$

$=$

$\sup_{t\in\mathbb{Z}_{\geq 2}}\frac{1}{n_t}(- \tfrac13(n_t+1) + n_t - \frac13(n_t+1)) $

$=$

$\sup_{t\in\mathbb{Z}_{\geq 2}}(\tfrac13 - \frac{2}{3n_t} )$

$=$

$\frac13$

the latter since arbitrarily large Andrásfai graphs exist.

Now let us write, for any natural number $k$,

$$q_k := \sup_{\text{all finite graphs $G$ with $\omega(G)=k$}}q_{\mathrm{relative}}(G) $$

for the quantity you are more intersted in.
A more important question than what value the single universal constant $q\in[\frac13,1]$ has, is to analyse the *function*

$$ S: \mathbb{N}\rightarrow [0,1] $$
$$ k\mapsto q_k$$.

It would be helpful for systematic reasons if others would use this notation.