The derived category of abelian groups is somewhat special that makes the Künneth and universal coefficient theorems take an unusually special form.
An abstract way to state this property is
Theorem: Every element of the derived category of abelian groups is the direct sum of one-term complexes
Proof: Every chain complex is quasi-isomorphic to a complex of free abelian groups. And if $C_\bullet$ is a complex of free abelian groups, the fact every subgroup of a free abelian group is free implies you can decompose $C_n = \ker(\partial_n) \oplus \mathrm{im}(\partial_n)$, and thus you can decompose $C_n$ into a direct sum of complexes of the form $\mathrm{im(\partial_n)} \to \ker(\partial_{n-1})$, each of which is isomorphic to the one-term complex $H_{n-1}(C_\bullet)$ concentrated in degree $n-1$. $\square$
In particular, the equivalence class of every chain complex $C_\bullet$ includes the complex
$$ \ldots \xrightarrow{0} H_1(C_\bullet) \xrightarrow{0} \underline{H_0(C_\bullet)} \xrightarrow{0} H_{-1}(C_\bullet) \xrightarrow{0} \ldots $$
which, of course, breaks apart into the direct sum of its individual terms.
From the form of tor and ext for one-term complexes, we can then write
$$ H_n (C_\bullet \otimes^\mathbb{L} D_\bullet)
\cong H_{n-i-j} \left( \bigoplus_i \bigoplus_j H_i(C_\bullet) \otimes^\mathbb{L} H_j(D_\bullet) \right)
\\ \cong
\bigoplus_i \bigoplus_j \begin{cases}
H_i(C_\bullet) \otimes H_j(D_\bullet) & n = i+j
\\ \mathrm{tor}(H_i(C_\bullet), H_j(D_\bullet)) & n = i+j +1
\end{cases}$$
The universal coefficient theorem is the special case where $C_\bullet$ is the complex of coefficients concentrated in degree zero. Similarly,
$$ H_n (\mathbb{R}{\hom}(C_\bullet, D_\bullet))
\cong H_{n+i-j} \left(\prod_i \bigoplus_j \mathbb{R}{\hom}(C_i, D_j)
\right)
\\ \cong
\prod_i \bigoplus_j \begin{cases}
\hom(H_i(C_\bullet), H_j(D_\bullet)) & n = j-i
\\ \mathrm{ext}(H_i(C_\bullet), H_j(D_\bullet)) & n = j-i-1
\end{cases}
\\\cong
\prod_i
\hom(H_i(C_\bullet), H_{n+i}(D_\bullet))
\oplus
\mathrm{ext}(H_i(C_\bullet), H_{n+i+1}(D_\bullet))
$$
With $D$ concentrated in degree zero, this becomes the familiar
$$ H_{-n} (\mathbb{R}{\hom}(C_\bullet, D))
\cong
\hom(H_n(C_\bullet), D)
\oplus
\mathrm{ext}(H_{n-1}(C_\bullet), D) $$