$\Phi$ is indeed an isomorphism. For injectivity, suppose
$$\Phi(m)=0$$
Then $\psi(m)=0$ and $g(m)=0$. $g(m)=0$ means that $m$ is in the image of $f$. Thus there is an $n\in N$ such that $f(n)=m$. Then $\psi(f(n))=n=0$, so we must have had that $n=0$, hence $m$ was $0$ in the first place.
For surjectivity, given $(n,l)$ we want to find $m$ such that $\psi(m)=n$ and $g(m)=l$. Take $m_1\in M$ so that $g(m_1)=l$. If we have $\psi(m_1)=n'$, take $m_1'=m_1-f(n')$; then $g(m_1')$ is still equal to $l$ because $f(n')\in\mathrm{Ker}(g)$, and $\psi(m_1-f(n'))=0$. Finally take $m_2\in\mathrm{Ker}(g)$ such that $\psi(m_2)=n$. Then $\Phi(m_1'+m_2)=(n,l)$.
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
"Why can we write...?":
If $b \in B$, then $b = b - \psi(\lambda(b)) + \psi(\lambda(b))$. Now we get $$\lambda(b - \psi(\lambda(b))) = \lambda(b) - \lambda(\psi(\lambda(b))) = \lambda(b) - \lambda(b) = 0,$$ such that $b -\psi(\lambda(b)) \in \text{ker}(\lambda)$. Therefore $b = b - \psi(\lambda(b)) + \psi(\lambda(b)) \in \text{ker}(\lambda) + \text{im}(\psi)$.
Why is that unique?:
Let $b = \psi(a) + \tilde{b} \in \text{ker}(\lambda) + \text{im}(\psi)$ and $b = \psi(a') + \tilde{b}' \in \text{ker}(\lambda) + \text{im}(\psi)$. Then we get $$\psi(a) - \psi(a') +\tilde{b} - \tilde{b'} = \psi(a) + \tilde{b} - (\psi(a') + \tilde{b}') = 0.$$ Applying $\lambda$ yields $a - a' = 0$ as $\tilde{b}, \tilde{b'} \in \text{ker}(\lambda)$, such that $a = a'$. Since $a = a'$, we get $\psi(a) = \psi(a')$ and therefore $$\tilde{b} - \tilde{b'} = \psi(a) - \psi(a') +\tilde{b} - \tilde{b'} = 0.$$ Thus we also have $\tilde{b} = \tilde{b'}$. This means we just showed uniqueness.
"Why is $\text{ker}(\lambda) \cong C$?":
Denote the restriction $\varphi \mid_{\text{ker}(\lambda)} \colon \text{ker}(\lambda) \rightarrow C$ by $\phi$. We will show that $\phi$ is an isomorphism. As $\varphi$ is surjective, any element $c \in C$ has a preimage $b = \psi(a) + \tilde{b}$ (we showed that every $b \in B$ can uniquely be written like that). That means we have $$\varphi(\tilde{b}) = \varphi(\psi(a)) + \varphi(\tilde{b}) = \varphi(\psi(a) + \tilde{b}) = c$$ by exactness. Thus $\phi$ is surjective as we found a preimage in $\text{ker}(\lambda)$. Now consider $b \in \text{ker}(\lambda)$, such that $\phi(b) = 0$, i.e. $b \in \text{ker}(\phi) \subset \text{im}(\psi)$. Thus it suffices to show that the intersection $\text{ker}(\lambda) \cap \text{im}(\psi)$ is trivial. Let $b \in \text{ker}(\lambda) \cap \text{im}(\psi)$. Then there exists $a \in A$, such that $\psi(a) = b$ and we also have $\lambda(b) = 0$. Thus $a = \lambda(\psi(a)) = \lambda(b) = 0$. This yields $b = 0$, such that $\phi$ is injective. This means $\phi$ is an isomorphism.