I know that there does not exist an isomorphism from $Z_8$ ⊕ $Z_2$ to $Z_4$ ⊕ $Z_4$ as there exist an element of order 8 in $Z_8$ ⊕ $Z_2$ and no element of order 8 in $Z_4$ ⊕ $Z_4$.
But what about homomorphisms. How many of them exist and how to find them.
There does exist homomorphisms. Example: if (1,0) of $Z_8$ ⊕ $Z_2$ is mapped to (1,0) of $Z_4$ ⊕ $Z_4$ and (0,1) of $Z_8$ ⊕ $Z_2$ mapped to (0,0) of $Z_4$ ⊕ $Z_4$.
The direct sum is a biproduct in the category of abelian groups. This means that is the coproduct and the product at the same time.
The coproduct $A\coprod B$ has the universal property that any map $f\colon A\coprod B\to C$ is uniquely defined by the two maps $f\circ i_1$ and $f\circ i_2$ where $i_1$ (resp. $i_2$) is the inclusion of $A$ (resp. $B$) in $A\coprod B$.
Using this you can show that $$ \text{Hom}\left(A\coprod B, C\right) \cong \text{Hom}(A, C) \times \text{Hom}(B, C) .$$
Dually, (it is the same with domain and codomain exchanged) a map into the product is uniquely determined by the composition with the projections of the product. So you have $$ \text{Hom}(A, B \times C) \cong \text{Hom}(A, B ) \times \text{Hom}(A, C) .$$
Now we know that $$ \text{Hom}(\mathbb{Z}/8\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}) \cong \text{Hom}(\mathbb{Z}/8\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}) \times \text{Hom}(\mathbb{Z}/8\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}) \times \text{Hom}( \mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}) \times \text{Hom}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}) .$$
Using $$ \text{Hom}(\mathbb{Z}/ n\mathbb{Z}, \mathbb{Z}/ m\mathbb{Z}) \cong \mathbb{Z} / gcd(n,m) \mathbb{Z} $$ it simplifies to $$ \text{Hom}(\mathbb{Z}/8\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}) \cong \mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z}/2 \mathbb{Z} .$$
Thus there are $2^6$ homomorphisms.