In Stacks, Section 13.9, one can find the definition of term-wise split exact sequences of complexes: we say that the exact sequence of complexes $ 0 \rightarrow A^\bullet \rightarrow B^\bullet \rightarrow C^\bullet \rightarrow 0 $ is term-wise split if $ B^\bullet = A^\bullet \oplus C^\bullet $ in a way that is compatible with the morphisms $ A^\bullet \rightarrow B^\bullet $ and $ B^\bullet \rightarrow C^\bullet $.
I understood it as saying that the complex $ 0 \rightarrow A^\bullet \rightarrow B^\bullet \rightarrow C^\bullet \rightarrow 0 $ is isomorphic (in $ \mathrm{Kom}(A) $) to $ 0 \rightarrow A^\bullet \rightarrow A^\bullet \oplus C^\bullet \rightarrow C^\bullet \rightarrow 0 $. However, it seems to me that this wouldn't be the case, since afterwards the connection morphism $ C^\bullet \overset\delta\rightarrow A[1]^\bullet $ is defined as the composite $$ C^\bullet \rightarrow A^\bullet \oplus C^\bullet \overset{d}\rightarrow A[1]^\bullet \oplus C[1]^\bullet \rightarrow A[1]^\bullet $$ But wouldn't that composite vanish given that $ d $ is just $ d_A \oplus d_B $?
What is actually meant by compatibility with those morphisms?
The definition of termwise split is merely that for each $n$, the short exact sequence $0\to A^n\to B^n\to C^n\to 0$ splits. This gives an isomorphism $B^n\cong A^n\oplus C^n$, but crucially, this splitting is not assumed to be compatible with the differentials in the chain complex. So, for instance, for each $n$ we have a map $C^n\to B^n$ which splits the surjection $B^n\to C^n$, but these maps $C^n\to B^n$ may not form a morphism of chain complexes $C^\bullet\to B^\bullet$. As a result, $B^\bullet$ is not necessarily isomorphic to $A^\bullet\oplus C^\bullet$ as a chain complex; we only have objectwise isomorphisms $B^n\to A^n\oplus C^n$ which may not commute with the differentials.