Prove that if $0 \to A \to B \to C \to 0$ is an exact sequence, with $f:A \to B$ and $g: B \to C$, all $R$-modules with $B = f(A) \oplus D$ for some $R$-module $D$, then there exists an $R$ homomorphism from $C$ to $B$ such that $gg' = Id_C$.
Here is my attempt: let $c \in C$. Then since $g$ is surjective, $c=g(b)$ for some $b \in B$. Write (uniquely) $b = f(a)+d$ for some $a \in A, d \in D$. Then we define $g'(c) :=f(a) +d$. It can easily be seen that $gg' = Id_C$ and $g'$ is a $R$-homomorphism.
But I am having trouble showing that this is well defined. Here is my attempt: suppose $c = g(f(a)+d) = g(f(a')+d')$. Then $g(f(a-a')+(d-d')) = 0$. Hence $f(a-a')+(d-d')$ is in $\ker g = $ Im$ f$, so that there is $x \in A$ such that $f(x) = f(a-a') +d-d' \in B = f(A) \oplus D$. By uniqueness of representation, $d=d'$ and $f(x) = f(a-a')$. Thus, by injectivity of $f$, $x=a-a'$. Hence (putting everything together) we have that $g'(c) = f(a') + f(x) + d = f(a') + d$. How can I show that $f(x) = 0$? In other words, How is it that $f(a) = f(a')$?
You're having trouble showing that $f(a) = f(a')$ because in fact it needn't be true! Since the composition $gf$ is zero, the fact that $f(a)$ and $f(a')$ map to the same place under $g$ is actually no condition at all.
On the bright side, this means that if you define $g'(c)$ to be just $d,$ you will still have $gg' = Id,$ and the rest of your argument should work.
By the way, the upshot (which you may already understand, based on your title) is that actually $B \cong A \oplus C.$ The map $f$ is including $A$ as the first summand and this $g'$ you're constructing is including $C$ as the second summand.