-as suggested after the discussion in the comments-

i am understanding that the OP is asking whether a linear map $j:G \to H$ is functorial, in the sense that the image of the dual map $j^*:H^* \to G^*$ preserves the finite dual hopf algebra $H^\circ$, i.e.
$
j^*(H^\circ)\subseteq G^\circ
$

The answer is generally no -at least for an arbitrary linear map $j$- and in my understanding this has already been shown in the counterexample proposed by user66288 in the comments to the OP.

However, it will be the case, i.e. $j$ will be functorial if it vanishes on an ideal of finite codimension in $G$, since then the image of the finite dual $H^\circ$ will be inside the finite dual $G^\circ$, i.e. the image $j^*(f)$ of an arbitrary element $f\in H^\circ$ will be vanishing on an ideal of finite codimension in $G$:
$
j^*(f)=f\circ j\in G^\circ
$
for all $f\in H^\circ$.

(Notice that in this situation we actually have that $j^*(f)\in G^\circ$
for all $f\in H^*$, and this is something which makes me wonder whether it is too restrictive, as i have already mentioned in the comments to the OP).

On the other hand, this is not an IFF statement, in the sense of the first of my comments above:

Consider an ideal $R$ of $G$ of finite codimension, not necessarily contained in the kernel of $j$. Then, those $f\in H^\circ$ for which $j(R)\subseteq ker f$ will be mapped in $G^\circ$ under $j^*$. If this happens for all $f\in H^\circ$ then $j^*(H^\circ)\subseteq G^\circ$, so we cannot necessarily conclude the converse statement:

"*if $j$ is functorial then it vanishes on an ideal of finite codimension in $G$*"

**Edit (Nov 17):** i was thinking that an iff characterization of the functoriality of $j$, is still possible, in the sense of the last paragraph above:

a linear map $j:G\to H$, between two hopf algebras $H$, $G$ is functorial if and only if for any $f\in H^\circ$, there is an ideal $R_f$ of finite codimension in $G$, which is mapped inside the kernel of $f$ under $j$, i.e. $j(R_f)\in ker f$

(I am not sure if this implies that $j$ should be an algebra map then).