The exact sequence of groups
$1 \to N \overset{\phi}{\longrightarrow} G \overset{\psi}{\longrightarrow} H \to 1$
is split if $(1)$ there is a right inverse for $\psi$.
I need to show the the following conditions are equivalent to $\psi$ having a right inverse:
$(2)$ There is a subgroup $L$ of $G$ such that $L$ maps isomorphically to $H$ via $\psi$.
$(3)$ There is a subgroup $L$ of $G$ such that $\phi(N) \cap L = \{ e \}$ and $\phi(N)L = G$
$(1) \Rightarrow (2)$ is easy. If there is a right inverse $\rho$ of $\psi$, then it is injective, so $H \cong \rho(G) = L$. We define $\psi': L \to H$ by setting $\psi'(l) = \psi(l)$.
$\psi'$ is surjective, since it has a right inverse $\rho': H \to L$ defined by $\rho'(h) = \rho(h)$. $\psi'$ is also injective, since $\psi'(\rho(h_1)) = \psi'(\rho(h_2)) \Leftrightarrow h_1 = h_2 \Rightarrow \rho(h_1) = \rho(h_2)$
What about other implications? If some of them are wrong, please, say which ones and help to prove the right ones.
I'll try to give some hints.
$(2)\Rightarrow(1)$. If $(2)$ holds then the restriction $\psi|_L \colon L \to H$ is an isomorphism. Try guess now what is the right inverse of $\psi$.
$(2)\Leftrightarrow(3)$. A subgroup $L$ verifies $(2)$ if and only if $\ker \psi \cap L = (0)$ and $\psi|_L$ is surjective (namely that for each $h \in H$ there is a $l \in L$ such that $\psi(l)=h$). Combine this stuff with the fact that $\ker \psi=\phi(N)$ and some of the results you have probably seen to prove the first homomorphism theorem, I believe you should be able to prove the wished result.
Hope this helps.
Some addenda: from the comment below it seems that there is need for some additional information about the bi-implication $(2) \Leftrightarrow (3)$.
By very basic results of group theory we know that $G = \coprod_{g \in R}g \ker \psi$ where $R$ is a set of representatives for the $g \ker \psi$ (i.e. the elements of the set $G/\ker \psi$). From this it follows that for every $g \in G$ there are a $r \in R$ and a $k \in \ker \psi$, uniquey determined, such that $g=rk$.
Now whenever we have a subgroup $L \leq G$ such that for each $h \in H$ there is a unique $l \in L$ such that $\psi(l)=h$, we have the equality $\psi^{-1}(\{h\})=l\ker \psi$. This implies that $L$ is also a set of representatives for $G/\ker \psi$ and so for the previous remark $$G=\coprod_{l \in L} l\ker \psi = L \ker \psi=L \phi(N)\ .$$