If by "element" they mean "global element", then a global element (of $A$) is an arrow from $1\to A$ where $1$ is the terminal object. Since $T$ is the top element of the poset, it is the terminal object of that poset when viewed as a category. Since $\mathsf{Hom}$ is continuous in its second argument, we have that $\mathsf{Hom}(-,T)\cong 1$ where $1$ is the terminal object of the category of presheaves. A global element of the presheaf $P$ is an element of $$\mathsf{Nat}(1,P)\cong\mathsf{Nat}(\mathsf{Hom}(-,T),P)\cong P(T)$$ where $\mathsf{Nat}(F,G)$ is the set of natural transformations between $F$ and $G$, i.e. the hom-set of the category of presheaves. The second isomorphism is Yoneda. So the global elements of the presheaf $\Omega$ are in correspondence with the elements of the set $\Omega(T)$.
More naturally, we can use the notion of the internal language of an elementary topos which can be presented as the Mitchell-Bénabou language. This lets us write what looks like "normal" mathematics but interpret it into any elementary topos. In this context, global elements correspond to closed terms. In other words, the elements of $\Omega(T)$, in this case, correspond to the closed terms of type $\Omega$ in the internal language.
Here's the idea.
The monomorphisms in the presheaf category are pointwise monomorphisms, so we can identify subobjects of a presheaf $X\in[\mathbf{A}^\text{op},\newcommand\Set{\mathbf{Set}}\Set]$ with subpresheaves of $X$, in the sense of presheaves $F$ such that $F(a)\subseteq X(a)$ for all objects $a$, and for $f:a\to a'$, $X(f)$ sends elements of $F(a')$ to $F(a)$.
Now suppose we have a subpresheaf $F$ of some presheaf $X$. We want to construct a natural transformation $X\to \newcommand\Sub{\operatorname{Sub}}\Sub(H_{-})$.
Thus for $a\in \newcommand\A{\mathbf{A}}\A$, $\alpha\in X(a)$, we need to construct a subpresheaf of $H_a$. By Yoneda, $\alpha$ corresponds to a natural transformation
$H_a\to X$, so we can just take the preimage of $F$ in $H_a$. In other words, define $G_\alpha\newcommand\into\hookrightarrow\into H_a$ by
$$G_\alpha(a') = \{ f : a'\to a \text{ such that } f^*\alpha \in F(a')\subseteq X(a')\}.$$
Then we define $\eta : X\to \Sub(H_-)$ by $\eta \alpha = G_\alpha$.
Conversely, from a natural transformation, $\eta : X\to \Sub(H_-)$, we can recover the subobject $F$ by
$$F(a) =\{ \alpha\in X(a) \text{ such that } 1_a \in (\eta_a\alpha)(a)\subseteq H_a(a)\}.$$
Side note: a subfunctor is a subobject of a functor, and a sieve is a subobject of a representable functor, but we don't really need to use these words to prove the claim.
Edit:
To see that $\eta$ is natural, let $f:a\to a'$, let $\alpha\in X(a')$.
We need to show that $\eta f^*\alpha = f^*\eta\alpha$.
Now
$$
(\eta f^*\alpha)(a'')
=
\{
g: a''\to a \text{ such that } g^*f^*\alpha \in F(a'')
\},
$$
and
$$
(f^*\eta\alpha)(a'')
=
(f_*)^{-1}((\eta\alpha)(a''))
=
\{
g: a''\to a \text{ such that } (f\circ g)^* \alpha \in F(a'')
\}.
$$
Thus, since $(f\circ g)^* = g^*f^*$, we have naturality.
Edit 2
I've been asked how we show that if we start with a natural transformation $\eta : X\to \Sub(H_-)$ and construct the associated subobject $F$ of $X$ how we show that the natural transformation $\overline{F}$ associated to $F$ is in fact $\eta$.
Let $a,a'\in \mathbf{A}$. Recall that
$$F(a) = \{ \alpha \in X(a) \text{ such that } 1_a \in \eta_a(\alpha)(a) \}.$$
We also know that if $\alpha \in X(a)$, then
$$
\overline{F}_a(\alpha)(a')
=
\{
g:a'\to a
\text{ such that }
g^*\alpha \in F(a)
\}.
$$
Putting these together we can compute
$$
\begin{aligned}
\overline{F}_a(\alpha)(a')
&=
\{
g:a'\to a
\text{ such that }
1_{a'}\in \eta_{a'}(g^*\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
1_{a'}\in g^*(\eta_{a}\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
1_{a'}\in (g_*)^{-1}(\eta_{a}\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
g\circ 1_{a'}\in (\eta_{a}\alpha)(a')
\}
\\
&=
\{
g:a'\to a
\text{ such that }
g\in (\eta_{a}\alpha)(a')
\}
\\
&=(\eta_a\alpha)(a').
\end{aligned}
$$
Thus as subobjects of $H_a$, we have that
$\eta_a\alpha = \overline{F}_a\alpha$, as desired.
Best Answer
Short version:
The definition of the subobject classifier can be succinctly stated as "$\Omega$ represents the subobject functor". Since the subobject functor is contravariant, the universal property of $\Omega$ must be given in terms of maps into $\Omega$.
Long version:
For any (well-powered, for size reasons) category $C$, and any object $X$ in $C$, write $\mathrm{Sub}(X)$ for the set of subobjects of $X$ (the monic arrows into $X$, up to isomorphism over $X$). We can turn $\text{Sub}$ into a functor by defining its action on morphisms. Here the most natural choice is to make $\text{Sub}$ contravariant, i.e. into a functor $C^{\text{op}}\to \mathsf{Set}$. Why? Well, given an arrow $f\colon X\to Y$ and a subobject $i\colon S\hookrightarrow Y$, we can pull back $i$ along $f$ to get a subobject $S\times_Y X\hookrightarrow X$. In the category $\mathsf{Set}$, this is exactly the preimage operation.
Now $C$ has a subobject classifier exactly when the functor $\text{Sub}$ is representable. This means there is an object $\Omega$ and a natural isomorphism $\text{Sub}(-)\cong \text{Hom}(-,\Omega)$. The fact that $\text{Sub}$ is a contravariant functor on $C$ means that we have to use the contravariant Hom functor $\text{Hom}(-,\Omega)$ (instead of the covariant Hom functor $\text{Hom}(\Omega,-)$), and thus subobjects of $X$ are classified by arrows $X\to \Omega$.
At this point you might ask if we could turn $\text{Sub}$ into a covariant functor instead. After all, given an arrow $f\colon X\to Y$ in $\mathsf{Set}$, there is a natural way to turn a subobject of $X$ into a subobject of $Y$: take its image along $f$. Well, this idea only works in a category with a natural notion of "images" of subobjects under arbitrary arrows, e.g. a regular category or an adhesive category. But indeed there are categories where we can make $\text{Sub}$ into a covariant functor.
However, even if we can make $\text{Sub}$ into a covariant functor, we should not typically expect it to be representable. This is because representable functors preserve limits, and the subobject functor rarely preserves limits.
For example, writing $\text{Sub}$ for the covariant subobject functor, if there is an object $\Omega$ such that $\text{Sub}(-)\cong \text{Hom}(\Omega,-)$, then for any product $X\times Y$ in $C$, we have $$\text{Sub}(X\times Y)\cong \text{Hom}(\Omega,X\times Y)\cong \text{Hom}(\Omega,X)\times \text{Hom}(\Omega,Y)\cong \text{Sub}(X)\times \text{Sub}(Y).$$ I hope it's clear that this property is not true in $\mathsf{Set}$ or any other familiar categories (the only natural categories I can think of in which this is true are preorders).
It's much more natural to ask the contravariant subobject functor to be representable. Here, preserving limits means turning colimits in $C$ (limits in $C^{\text{op}}$) into limits in $\mathsf{Set}$. So for example we should have $\text{Sub}(X\sqcup Y)\cong \text{Sub}(X)\times \text{Sub}(Y)$. In $\mathsf{Set}$, this says that picking a subobject of a disjoint union of $X$ and $Y$ is the same as picking a subobject of $X$ and a subobject of $Y$.