Suppose you face a single buyer whose willingness to pay, $v$, is distributed according to $F(v)$. If you charge a price $p$, he will buy if and only if $v>p$, leaving you with expected revenue of $$r(p)=\Pr(v>p)p=[1-F(p)]p.$$
Let's maximise revenue by computing an FOC:
$$r'(p)=1-F(p)-F'(p)p=0.$$
We can rearrange this as
$$\phi(p)\equiv p-\frac{1-F(p)}{F'(p)}=0$$
i.e., the 'virtual valuation' should be zero.
If we return to $r'(p)$ and think about what the individal terms mean, we can see where the "cost" you ask about comes from: a unit increase in price causes an extra unit of revenue in the fraciton $1-F$ of the time that the buyer is willing to buy, but reduces the liklihood of him buying by $F'$. This is the fundamental trade-off that a seller who doesn't know the buyer's willingness to pay must make.
If you know anything about standard monopoly theory then this setup should be quite familiar. Usually, when we look at a profit-maximising monopolist with demand $D(p)$ and zero marginal cost we solve
$$\max_p D(p)p\iff \underbrace{D'(p)p-D(p)}_{\text{marginal revenue}}=0.$$
In an auction-like setting, the 'demand' $1-F(p)$ is simply the probability of the buyer being willing to pay $p$.