I was wondering if anyone could help with how to, strictly from Yoneda's Lemma, obtain the coproduct map on the Hopf Algebra for an Affine Group Scheme. Particularly for something like $\text{SL}_2$
So if $G=\text{SL}_2$, let $m:G \times G \to G$ be the multiplication map. Yoneda'e lemma tells us that this induces a map $\Delta:A \to A \otimes A$ and there is a specific construction:
$m: G\times G(A \otimes A) \to G(A \otimes A)$ which (since $G$ is representable is just a map $\text{Hom}(A \otimes A, A \otimes A) \to \text{Hom}(A, A \otimes A)$ and element-wise $id$ gets sent to $\Delta$.
This is hypothetically how you should "find" $\Delta$. In the case of $\text{SL}_2$ I guess the part I am having trouble with is finding the corresponding element in $G\times G(A \otimes A)$ that goes with $id$. That way I can just do the multiplication map and see what the entries are to get the $\Delta$ map.
How about we do $\mathbb G_m$ (or $GL_1$, if you prefer to denote it that way),so there is less notation.
The ring $A$ is equal to $k[x,x^{-1}]$, so $A \otimes A $ equals $k[x,x^{-1},y,y^{-1}]$ (thinking of $x$ as $x\otimes 1$ and $y$ as $1 \otimes x$, if you see what I mean).
Now the identity map from $A$ to $A$ corresponds to the element $x \in \mathbb G_m(A)$ (since the identification of $\mathrm{Hom}(A,A')$ with $\mathbb G_m$ is given by sending a homomorphism $\varphi$ to the element $\varphi(x) \in A'$), and so the identity map of $A\otimes A$ corresponds to the pair $(x,y) \in (\mathbb G_m\times \mathbb G_m)(A\otimes A) = \mathbb G_m(A\otimes A) \times \mathbb G_m(A\otimes A)$.
When we multiply these two elements, we get $xy \in \mathbb G_m(A\otimes A)$, and thus $\Delta$ is given by $x \mapsto xy$.
In general, $\Delta$ just expresses the formula for $m$ in coordinates. So for $SL_2$, if we label the generators of $A$ as $a,b,c,d$ (with the relation $ad - bc = 1$), and label the generators of $A \otimes A$ as $a,b,c,d,a',b',c',d'$ (rather than $a\otimes 1$, etc., $1\otimes a$, etc.), then $\Delta$ will have the formula $$a \mapsto a a' + bc', \text{ etc.},$$ just coming from the formula $$\pmatrix{a & b \\ c & d}\cdot \pmatrix{a' & b'\\ c' & d'} = \pmatrix{a a' + b c' & a b' + b d' \\ c a' + d c' & c b' + d d'}.$$
If you want to derive this via your Yoneda strategy, it will work exactly as with the $\mathbb G_m$ example (just with more notation, because there are more generators for $A$).