I would claim that the splitting (and indeed the whole universal coefficient theorem) is not really a topological theorem. If we take the homological version one really works with the chain complex $C_*(X)$$C_\ast(X)$ in the derived category of $\mathbb Z$-complexes. We then have $C_*(X,M)=C_*(X)\bigotimes M$$C_\ast(X,M)=C_\ast(X)\bigotimes M$ but as $C_*(X)$$C_\ast(X)$ is free this equals the derived tensor product $C_*(X)\bigotimes^{\mathbb L} M$ and hence is a formula in the derived category. One can then use the fact that in the derived category of $\mathbb Z$-modules every complex is isomorphic to the sum of its (shifted) homology: $C\cong \bigoplus_nH_n(C)[n]$ so that $$ C_*(X)\bigotimes^{\mathbb L} M \cong \bigoplus_n(H_n(X)\bigotimes^{\mathbb L} M)[n] $$ and as $A\bigotimes^{\mathbb L} M\cong A\bigotimes M\bigoplus \mathrm{Tor}^1(A,M)[1]$ we get the universal coefficient formula including the splitting.
This idea also demonstrates why the splitting is not canonical. We may for instance consider a group $G$ acting on $X$. We then get at complex $C_*(X)$ in the derived category of $G$-modules and a complex in that category is in general not isomorphic to the sum of its homology.
On the other hand, this technique can be used (essentially) each time some invariant of a topological space $X$ only depends on its chain complex in a way that takes quasi-isomorphisms to isomorphisms. The conclusion is that it only depends on the homology of $X$. A nice example is the homology of the $n$'th symmetric product of $X$. It turns out to be the homology of a complex constructed functorially from $C_*(X)$ and exactly in a way that preserves quasi-isomorphisms. Hence it only depends on the homology of $X$ (and one can also give explicit formulas).
However, the method that you declare a fondness for is also useful if one goes beyond homology. It can be used to give a universal coefficient spectral sequence (due to Adams I think) for the (co)homology with coefficients in module spectrum over a ring spectrum. In general this spectral sequence does not degenerate to short exact sequences so the problem of splitting is not even (that) relevant. However, for for instance $K$-theory it does but I imagine (though I don't know but others certainly do) that even there one can find examples of non-splitting.