Let $$0 \rightarrow M_1 \rightarrow N \rightarrow M_2 \rightarrow 0$$ be an exact sequence of $R$-modules. Suppose there is any $R$-module homomorphism $g:N \rightarrow M_1 \oplus M_2$ making the diagram$$\require{AMScd} \begin{CD} 0 @>>> M_1 @>\phi_1>> N @>\phi_2>> M_2 @>>> 0 \\ @. @| @VgVV @| @. \\ 0 @>>> M_1 @>\iota>> M_1 \oplus M_2 @>\pi>> M_2 @>>> 0 \end{CD} $$ commute. $\iota$, $\pi$ are the inclusion and identity maps. Prove that the sequence above splits.
Attempt to prove surjectivity: Let $(a,b) \in M_1 \oplus M_2$. Then $(a,b) \mapsto b$ in $M_2$. Since $\phi_2$ is surjective, there exists $c \in N$ such that $\phi_2(c)=b$. By commutativity, $\text{id} \circ \phi_2=\text{id} \circ g$. So that $\pi(g(c))=b$ and $g(c)=(d,b)$ for some $d \in M_1$. Since $a \in M_1$, by commutativity $g \circ \phi_1=\iota \circ \text{id}$ and $g(\phi_1(a))=\iota \circ \text{id}(a)=(a,0)$. Similarly, $g(\phi_2(d))=\iota \circ \text{id}(d)=(d,0)$. Then $g(\phi_1(a)+c-\phi_2(d))=(a,0)+(d,b)-(d,0)=(a,b)$
I'm not sure what definition of split-exact you are working with, but the definition you will find on Wikipedia is basically the existence of the above commutative diagram, with $g$ being an isomorphism.
Let me show you how to argue that $g$ is injective, and you can show surjectivity yourself. Suppose $g(x) = 0$ for $x \in N$. Then by commutatitivity of the diagram, $\phi_{2}(x) = \pi(g(x)) = \pi(0) = 0$. Thus $x \in \operatorname{ker} \phi_2$, and so by exactness, is in the image of $\phi_1$, say $x = \phi_1(y)$ for $y \in M_1$. Again by commutativity, $g(x) = g(\phi_1(y))$ is nothing but $y \oplus 0$. If this is zero, that means that $y=0$, so $x = \phi_{1}(0) = 0$, and injectivity is proven.