For a short exact sequence $0 \longrightarrow A \overset{i}{\longrightarrow} B\overset{j}{\longrightarrow} C \longrightarrow 0$ of abelian groups the following statements are equivalent:
There is a homomorphism $p : B\to A$ such that $p \circ i = \text{id}_A$.
There is a homomorphism $s : C \to B$ such that $j \circ s = \text{id}_C$
There is an isomorphism $B \cong A \oplus C$.
I am trying to prove that $1 \implies 3$ and $2 \implies 3$ and afterwards that $3 \implies 1$ and $3 \implies 2$.
I'm stuck in the first part i.e. $1 \implies 3$. Suppose that $1$ is true, then there exists $p : B \to A$ such that $p \circ i = \text{id}_A$. I now tried to define $$\varphi : B \to A \oplus C, \ b \mapsto (p(b), j(b))$$ and an inverse as $$\varphi^{-1}: A \oplus C \to B, (a,c) \mapsto i(a).$$ What I got is that $$\varphi(\varphi^{-1}(a,c)) = \varphi(i(a))= (p(i(a)), j(i(a))) = (a, j(i(a))) \ne \text{id}_{A \oplus B}$$ and that $$\varphi^{-1}(\varphi(b))=\varphi^{-1}(p(b), j(b)) = i(p(b)) \ne \text{id}_B.$$
So I think that I've defined the inverse wrong here, but I couldn't think of any other candidates for it. Is there some other way to define it?
In addition to the need to strengthen condition (3) to make sure it is equivalent, your proposed inverse for $\varphi$ cannot work, since it has nontrivial kernel given by $\{0\}\oplus C$. Instead, some more work is needed to use $p$ to establish the implication $(1)\Rightarrow(3)$.
Given any $c\in C$, let $b_c$ be an arbitrary element of $B$ such that $j(b_c)=c$. now consider $b=b_c-i(p(b_c))$.
First, note that $j(b) = j(b_c) - j(i(p(b_c)) = j(b_c)=c$. And $p(b) = p(b_c)-p(i(p(b_c))) = p(b_c) - p(b_c) = 0$.
Second: suppose $r_c\in B$ is another element of $B$ with $j(r_c)=c$. Let $r=r_c-i(p(r_c))$. Then $j(b-r) = j(b)-j(r) = c-c = 0$, and since the original sequence is exact, we have that $b-r\in \ker(j) =\mathrm{im}(i)$. Thus, there exists $a\in A$ such that $i(a)=b-r$. But then $$ a = p(i(a)) = p(b-r) = p(b)-p(r) = 0-0 = 0.$$ Thus, in fact we have $b=r$.
That means that even though the element $b_c$ is not uniquely determined by the fact that it is a pre-image of $c$, the map $p$ does provide a "canonical" choice of pre-image of $c$: namely, $x=b_c-i(p(b_c))$ for any pre-image $b_c$.
(Intuitively what is going on is that if $B$ "really is" $A\oplus C$, and the map $p$ is the projection onto the first coordinate, we are taking an arbitrary preimage $(a,c)$ of $c$, and then subtracting $i(p(a,c)) = i(a) = (a,0)$, so that we get the "canonical" pre-image $(0,c)$...)
So define $f\colon A\oplus C\to B$ by $$f(a,c) = i(a) + \Bigl(b_c - i(p(b_c))\Bigr),$$ where $b_c$ is any pre-image of $c$. This is well-defined by the work done above. Now we have $$\begin{align*} \varphi(f(a,c)) &= \varphi\bigl(i(a)+b_c-i(p(b_c))\bigr)\\ &= \bigl( p(i(a))+p(b_c-i(p(b_c))), j(i(a)) + j(b_c-i(p(b_c)))\bigr)\\ &= ( a+0,0+c) = (a,c)\\ f(\varphi(x)) &= f(p(x),j(x))\\ &= i(p(x)) + x - i(p(x)) \qquad\text{since }x\text{ is a pre-image of }j(x)\\ &= x. \end{align*}$$ So $f$ is the inverse of $\varphi$, proving that $\varphi$ is a bijection, as required.