You're working too hard. I'll assume you're convinced that $\mathbb{Z}$ is the free group on one generator, which means you're convinced that
$$\text{Hom}(\mathbb{Z}, G) \cong G$$
(where the RHS should really be the underlying set of $G$ but I'm omitting that I'm applying the underlying set functor). One way to state the universal property of the coproduct is that
$$\text{Hom}(X \sqcup Y, Z) \cong \text{Hom}(X, Z) \times \text{Hom}(Y, Z)$$
from which it follows that
$$\text{Hom}(\mathbb{Z} \sqcup \mathbb{Z}, G) \cong G \times G$$
and this is also the universal property of the free group on two generators, so you're done by the Yoneda lemma. The argument is exactly the same for the free group on $n$ generators, and in fact on a set's worth of generators.
More abstractly, the forgetful functor $\text{Grp} \to \text{Set}$ has a left adjoint, the free group functor. As a left adjoint, it preserves colimits, and in particular coproducts. But every set $X$ is the coproduct of $X$ copies of the $1$-element set, so it follows that the free group $F_X$ is the coproduct of $X$ copies of $\mathbb{Z}$.
I'd like to start with your first question about "almost kernel". The following construction can be brought to the case of "almost cokernel" without much modification.
Suppose $a:A\to M$ is an "almost kernel" and $k$ is the kernel map. By definition of $A$, $\varphi \circ a=0$ and there exists a map (not necessarily unique) $i:\text{Ker }\varphi \to A$ makes the following diagram commute:
$$\require{AMScd}
\begin{CD}
@. \text{Ker } \varphi @>i>> A @.\\
@. @| @VVaV \\
0 @>>> \text{Ker } \varphi @>k>> M @>\varphi>> N
\end{CD}$$
On the other hand, by the universal property of kernel there exists an unique map $r:A\to \text{Ker }\varphi$ makes the following diagram commute:
$$\require{AMScd}
\begin{CD}
A @>r>> \text{Ker } \varphi @.\\
@| @VVkV \\
A @>a>> M @>\varphi>> N
\end{CD}$$
We obtain another commutative diagram:
$$\require{AMScd}
\begin{CD}
@. \text{Ker } \varphi @>r\circ i>> \text{Ker }\varphi @.\\
@. @| @VVkV \\
0 @>>> \text{Ker } \varphi @>k>> M @>\varphi>> N
\end{CD}$$
Now by the universal property of kernel again $r\circ i=\text{id}_{\text{Ker}\varphi}$, since the identity also makes the above diagram commute. You can check that the existence of such $r,i$ is, in fact, the necessary and sufficient condition of an almost kernel. Ones can take $i$ to be the inclusion, then $r$ is called a retract of $A$ to $\text{Ker }\varphi$.
Such examples come from direct sums, since the composition of the projection from $M\oplus N$ onto $M$ and the obvious inclusion $M\to M\oplus N$ is the identity on $M$. In this case, you can take $\varphi: \mathbb{Z}\to 0$, then the kernel is $\text{id}:\mathbb{Z}\to \mathbb{Z}$ and an almost kernel is $\text{pr}_1: \mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}, (x,y)\mapsto x$. It is easy to see that $r=\text{pr}_1$ and a choice for $i:\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$ is $x\mapsto (x,0)$. Another choice is $i':x\mapsto (x,x)$.
In the above argument, the use of universal property is essential. The main purpose of universal property is to obtain the uniqueness of the kernel (or cokernel). Suppose that there are two kernel $k:K\to M$ and $k':K'\to M$ of $\varphi:M\to N$, then a similar argument to the one we presented above show that there exists unique $i:K\to K'$ and $j:K'\to K$ such that $k'=k\circ j,k=k'\circ i$ and $i\circ j=\text{id}_{K'},j\circ i=\text{id}_K$. The last two equations say that $K$ and $K'$ are isomorphic, and the previous two equations say that the isomorphisms are natural.
The second advantage of the universal property which is not very clear at the moment is that universal property provides morphisms. This is the case with tensor product, and you will use this frequently when deals with diagrams ("diagram chasing" is the name of this method). Further, the notion of kernel and cokernel is important because they are the building blocks of abelian category. On an abelian category, we can do homological algebra, a ubiquitous computational tool of algebraists.
I think what I wrote above will give you some grasp about how universality work. From your question, I think you have the same problem as mine when I first learn about category theory. The definition of kernel suggests that there are many kernels, but in every application of universal property they treat them like there is only one kernel. From the categorical viewpoint, two objects defined by the same universal property are identical. When there is a kernel, its universal property allows one to only care about the unique factorization. This is quite troublesome at first, but it is indeed an advantage of category theory.
Best Answer
So, first off, $\pi$ is going to be the obvious projection $g \mapsto gN$.
Now, we just need to check initiality. Let $\alpha: G' \to L$ be such that $\alpha \circ \varphi = 0$. Then we need a map $\beta: G'/N \to L$ such that $\alpha = \beta \circ \pi$. There's an obvious choice for this: define $\beta(gN) = \alpha(g)$. We need to check that it's well-defined: if $gN = hN$, then $gh^{-1}\in N$.
Suppose that $\alpha(g)\neq \alpha(h)$. then $gh^{-1} \not\in \ker(\alpha)$. But $\ker\alpha$ is a normal subgroup of $G'$, and since $\alpha\circ\varphi = 0$, $\ker\alpha\supseteq\mathrm{im}\varphi$, so $\ker\alpha\supseteq N\ni gh^{-1}$, a contradiction. Thus, $\alpha(g) = \alpha(h)$, and $\beta$ is well-defined.
Finally, uniqueness of $\beta$ is clear, so we're done.
Essentially, the answer to your question is "yes".
For your side-question, it means that for all such $\alpha$, there are maps $\delta: G' \to G'/N$ and $\varepsilon: G'/N \to L$ such that $\alpha = \varepsilon\delta$.