In his book "Algebraic Topology", Hatcher introduces the functor $Ext$ and claims it satisfies: $$ Ext(\mathbb{Z}_n,G)=G/nG $$ His proof is as follows (irrelevant parts are trimmed):
...comes from dualizing the free resolution: $$0 \rightarrow \mathbb{Z} \overset{n}{\rightarrow} \mathbb{Z} \rightarrow \mathbb{Z}_n \rightarrow 0$$ to produce the exact sequence:
I don't see how does dualizing the free-resolution produces the exact sequence.
Help will be greatly appreciated.
This is an exact sequence: $$ \mathbb Z \overset{n}{\to} \mathbb Z \to \mathbb Z_n \to 0$$
Applying the (contravariant) functor $\text{Hom}( \_, G)$ to this exact sequence, we get another exact sequence: $$ 0 \to \text{Hom}( \mathbb Z_n, G) \to \text{Hom}( \mathbb Z, G) \overset{n}{\to} \text{Hom}( \mathbb Z, G) \ \ \ \ \ \ (\star) $$
[It is a standard fact in homological algebra that if $L\to M \to N \to 0$ is exact, then $ 0 \to \text{Hom}( N, G) \to \text{Hom}( M, G) \to \text{Hom}( L, G) $ is exact.]
The following is a free resolution of $\mathbb Z_n$: $$ \dots \to 0 \to 0 \to \mathbb Z \overset{n}{\to} \mathbb Z $$
Applying the contravariant functor $\text{Hom}( \_, G)$ to this gives us a chain complex:
$$ \text{Hom}( \mathbb Z, G) \overset{n}{\to} \text{Hom}( \mathbb Z, G) \to 0 \to 0 \to \dots $$
I'll emphasise that this sequence we've just produced is not exact - it is merely a chain complex.
By definition, $\text{Ext}(\mathbb Z, G)$ is the quotient $$ \text{Ext}(\mathbb Z, G) = \frac{\text{Ker}[\text{Hom}( \mathbb Z, G) \to 0]}{\text{Im}[\text{Hom}( \mathbb Z, G) \overset{n}{\to} \text{Hom}( \mathbb Z, G)]}.$$
Clearly, $\text{Ker}[\text{Hom}( \mathbb Z, G) \to 0] = \text{Hom}( \mathbb Z, G) $, so $$ \text{Ext}(\mathbb Z, G) = \frac{\text{Hom}( \mathbb Z, G)}{\text{Im}[\text{Hom}( \mathbb Z, G) \overset{n}{\to} \text{Hom}( \mathbb Z, G)]}.$$
Thus the following is an exact sequence: $$ \text{Hom}( \mathbb Z, G) \overset{n}{\to} \text{Hom}( \mathbb Z, G) \to \text{Ext}(\mathbb Z, G) \to 0 \ \ \ \ \ \ (\star \star)$$
Combining $(\star)$ and $(\star\star)$, we see that the following sequence is exact: $$ 0 \to \text{Hom}( \mathbb Z_n, G) \to \text{Hom}( \mathbb Z, G) \overset{n}{\to} \text{Hom}( \mathbb Z, G) \to \text{Ext}(\mathbb Z, G) \to 0$$
This sequence is exact at $\text{Hom}( \mathbb Z_n, G)$ and at the first $\text{Hom}( \mathbb Z, G) $ since $(\star)$ is exact.
This sequence is exact at the second $\text{Hom}( \mathbb Z, G) $ and at $\text{Ext}(\mathbb Z, G) $ since $(\star\star)$ is exact.
Edit: I noticed that the title of your question is "Why does $\text{Ext}(\mathbb Z, G) = G / nG$?" If this is all your care about, then we don't need to do all this work to derive the exact sequence above. All we need to do is look at the formula: $$ \text{Ext}(\mathbb Z, G) = \frac{\text{Hom}( \mathbb Z, G)}{\text{Im}[\text{Hom}( \mathbb Z, G) \overset{n}{\to} \text{Hom}( \mathbb Z, G)]}.$$
The numerator of the quotient is $G$ and the denominator of the quotient is $nG$, so $\text{Ext}(\mathbb Z, G) = G / nG$.