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.