What would be a counterexample to the condition in the definition of a homomorphism that goes like down below?
$h(f^\mathfrak{A}(a_1,...,a_n)) = h(f^\mathfrak{B}(a_1,...,a_n))$
[as opposed to the original condition, $h(f^\mathfrak{A}(a_1,...,a_n)) = f^\mathfrak{B}(h(a_1),...,h(a_n))$]
Your proposed definition doesn't even "parse" in the first place: $h$ only applies to elements of $\mathfrak{A}$ and $f^\mathfrak{B}$ only applies to (and only outputs) elements of $\mathfrak{B}$, so there's no way to make sense of the right hand side expression "$h(f^\mathfrak{B}(a_1,...,a_n))$."
(OK, technically if $\mathfrak{A}$ and $\mathfrak{B}$ have "overlapping domains" (e.g. if $\mathfrak{A}=\mathfrak{B}$) this may happen to parse, but it won't make sense in general. And even in this case it won't behave well: for instance, as Troposphere observes every map $\mathfrak{A}\rightarrow\mathfrak{A}$ whatsoever would constitute a homomorphism according to your definition in the case $\mathfrak{A}=\mathfrak{B}$.)
