Let $C$ and $C'$ be two $\mathbf{K}$-algebras, and let $A$ and $A'$ be two $\mathbf{K}$-algebras. Let $\gamma\colon C\to C'$ be a $\mathbf{K}$-algebra morphism. Let $\alpha\colon A\to A'$ be a $\mathbf{K}$-algebra morphism. The map \begin{equation} \operatorname{Hom}(C',A)\to\operatorname{Hom}(C,A')\,,\,f\mapsto\alpha\circ f\circ\gamma \end{equation} is a $\mathbf{K}$-algebra homomorphism from the convolution algebras $((\operatorname{Hom}(C',A),*)$ and $((\operatorname{Hom}(C,A'),*))$.
Here is my attempt to solve the problem:
I did not look at the attempt. If you assume that $C$ and $C'$ are coalgebras, and that $\gamma$ is a coalgebra morphism, it is plain (that is what the notation and terminology suggests - and maybe the poster wanted all objects to be Hopf algebras in the first place as the tag suggests): $$ \alpha(f*g)\gamma(x)=\alpha(f(\gamma(x)_{(1)})g(\gamma(x)_{(2)})=\alpha f\gamma(x_{(1)})\alpha g\gamma(x_{(2)})=\alpha f\gamma*\alpha g\gamma(x)\,. $$