The proof of the Dold-Kan theorem basically amounts to the following.  Let $\mathbb{Z}\Delta$ denote the pre-additive category generated by the simplicial indexing category, so that $s\mathrm{Ab}=\mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})$, the category of additive functors.

Let $\mathcal{C}$ be the "Karoubi envelope" of $\mathbb{Z}\Delta$, the category obtained by freely adjoining splittings of idempotents.  You can identify the Karoubi envelope with a full subcategory $s\mathrm{Ab}$, namely the closure of the image of the Yoneda embedding under splitting idempotents.

Then it is trivial that $\mathrm{Fun}^{\mathrm{add}}(\mathcal{C}^\mathrm{op},\mathrm{Ab}) \approx \mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})=s\mathrm{Ab}$.  The Dold-Kan theorem amounts to observing that every object in $\mathcal{C}$ is a direct sum of objects $G(n)$, and that the full subcategory of $G(n)$s is the "indexing category" for chain complexes.  You also have that $G(n)\approx \mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$.

The usual formulas for normalized chains arise by explicitly identifying $G(n)$ with either a (split) subobject or a (split) quotient object of $\mathbb{Z}\Delta[n]$.  In fact, there is only one way (up to sign) to identify $G(n)$ as a split subobject of $\mathbb{Z}\Delta[n]$, and this translates into the "quotient by degeneracies" construction of normalized chains.  There are several ways to identify $G(n)$ as a split quotient: the usual presentation of normalized chains as subobjects is induced by the projection $\Delta[n]\to\Lambda^0[n]$.  

Of course, you could also be perverse: $G(n)$ is a summand of $\mathbb{Z}\Delta[k]$ for any $k\geq n$, so you could define normalized chains using your favorite idenification of $G(n)$ as a summand of $\mathbb{Z}\Delta[n+42]$, if you want!

**Added.**  Let me try to answer Karol's question, by giving a proof which doesn't use much computation.  

Set $F(n):=\mathbb{Z}\Delta[n]$ and  $G(n):=\mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$. The key observation is to show that the projection $F(n)\to G(n)$ admits a section.  I don't have a clever way to do that ... you just have to write down the formula.  (Which is: send the "generator" $\langle0,1,\dots,n\rangle$ of $G(n)$ to $\sum \pm\langle 0,a_1,\dots,a_n\rangle$, where the sum is over sequences with $a_k\in \{k-1,k\}$, and the sign is the parity of $\lvert\{k\,|\,a_k\neq k\}\rvert$.)  The nice thing is that the section turns out to be unique (this is hardly obvious to begin with).

Now filter $\Delta[n]$ by subcomplexes $\Delta^k[n]=$ the union of all $k$-simplices which contain the vertex $0$.  It's easy to see that
$$
\mathbb{Z}\Delta^k[n] /\mathbb{Z}\Delta^{k-1}[n] \approx G(k)^{\oplus \binom{n}{k}}.
$$
But we just showed that the $G(k)$ are *projective*, so the filtration lifts to a direct sum decomposition $F(n)\approx \bigoplus G(n)^{\binom{n}{k}}$.  

Now you want to compute $\hom(G(m), G(n))$.  Here is a trick.  It's clear that $\hom(F(m),F(n))=\mathbb{Z}^{\binom{n+m+1}{m+1}}$.  Since the $G(k)$ are summands of the $F(n)$, we have $\hom(G(i),G(j))\approx \mathbb{Z}^{a_{i,j}}$ for some $a_{i,j}\geq0$.  We want to show $a_{i,i}=a_{i,i+1}=1$ and all other $a_{i,j}=0$. 

It's easy to exhibit that $a_{i,i}\geq1$ and $a_{i,i+1}\geq1$.  Plugging the direct sum decomposition into $\hom(F(m),F(n))$ gives 
$$
\binom{n+m+1}{m+1} = \sum_{i,j} \binom{m}{i}\binom{n}{j} a_{i,j}.
$$
But you can use standard combinatorial identities to show this is *already* an equality when we set $a_{i,i}=a_{i,i+1}=1$ and other $a_{i,j}=0$.  So that must be the answer.