To address your first comment: you can be less imprecise there by being more explicit about the isomorphism. In particular, if the sequence $$
0 \to \ker \varphi \xrightarrow{i} M \xrightarrow{\varphi} N \to 0
$$
splits you have an isomorphism $f : M \to \ker \varphi \oplus N$ such that $f \circ i = \iota_1$ and $\pi_2 \circ f = \varphi$, where $\iota_1 : \ker \varphi \to \ker \varphi \oplus N$ and $\pi_2 : \ker \varphi \oplus N \to N$ are the corresponding inclusion and projection.
Then you have for $\iota_2 : N \to \ker \varphi \oplus N$ that $\pi_2 \circ i_2 = \mathrm{Id}_N$ and you define $r : N \to M$ by $r = f^{-1} \circ \iota_2$. Then $\varphi \circ r = \varphi \circ f^{-1} \circ \iota_2 = \pi_2 \circ \iota_2 = \mathrm{Id}_N$ so you have a right inverse.
For the converse, assume you have a right inverse $r : N \to M$. Note that a right inverse is injective, since $r(x) = r(y)$ implies $x = \varphi(r(x)) = \varphi(r(y)) = y$. Thus, we can identify $N$ with the submodule $\operatorname{im} r$ of $M$.
Our goal is to show that $M = \ker \varphi \oplus N$.
Now, to see how this can be done, let $m \in M$ . Then $n = \varphi(m)$ is an element of $N$ and since $r$ is a right inverse, for $m' = r(n)$ you have that $$\varphi(m - m') = \varphi(m) - (\varphi \circ r \circ \varphi)(m) = \varphi(m) - \varphi(m) = 0.$$
Thus $m - m' = i(k)$ for some unique $k \in \ker \varphi$ and we can write $m = i(k) + r(n)$.
Define $f : M \to \ker \varphi \oplus N$ by this decomposition, i.e. $$
f(m) = (k, \varphi(m)).
$$
You can verify this is an isomorphism with inverse $g : \ker \varphi \oplus N$ given by $g(k, n) = i(k) + r(n)$
Now, since $\varphi \circ i = 0$ you have $(f \circ i)(k) = (k, 0)$, i.e. is the inclusion $\iota_1 : \ker \varphi \to \ker \varphi \oplus N$. Similarly for $\pi_2 : \ker \varphi \oplus N \to N$ you have $(\pi_2 \circ f)(m) = \varphi(m)$, so $\pi_2 \circ f = \varphi$.
Thus the exact sequence splits, with $f$ giving the required isomorphism between exact sequences.
Best Answer
I think you can prove something stronger:
Let be $I$ and $J$ ideals of a ring $R$ ($R$-modules). Then the below sequence is exact: $$ 0 \to I \cap J \xrightarrow{f} I \oplus J \xrightarrow{g} I+J \to 0 $$
Indeed, you can identify $I \oplus J$ with $I \times J$. Then, you pick $$ f: I \cap J \to I \oplus J \qquad x \mapsto (x,x) $$
and
$$ g: I \oplus J \to I+J \qquad (x,y) \mapsto x-y $$
The first function is invective obviously, while the second is surjective (nothing to say I think). Moreover, if $a \in Im (F)$, then there exists $x \in I \cap J$ such that $a = (x,x)$. Then $g(a) = g(x,x) = x-x=0$, then $a \in \ker (g)$. Viceversa, if $(x,y) \in \ker(g)$ then $x = y$. Thus $(x,y) \in Im(f)$.
Clearly, your exercise is a particular case of what I've just said (vector spaces over a field $K$ are just $K$-modules)