To prove Proposition 5, he chooses to let $u = 1$, $u_1 = n$, $v = v_1 = g^{-1}$. I understood that u and u_1 are in the same coset N, and v and v_1 are trivially in the same coset. What I don't get is how these representative of special coset such N and another special case of $v = v_1$ could be used as the proof of a more general case i.e. $uvN = u_1 v_1 N$.
The statement in question is showing that $gng^{-1}\in N$, where $g$ and $n$ are general elements.
There is no problem with using other specific elements to obtain this proof.
In this part of the proof we know that $uvN=u_1v_1N$ for all elements satisfying the given condition. Therefore we know this result for any particular case and for a proof of this type you just have to think of a good special case to use.
"Short" answer:
In the proof part (1), there are actually two proofs because they are proving an "if and only if" statement.
In the first paragraph, starting with "Assume first that this operation is well-defined" through "... gives $gng^{-1}=n_1 \in N$, as claimed" is the proof of one direction.
The next part, starting with "Conversely, assume ..." and ending at the end of part (1) with "This proves that the operation is well defined" is a separate proof.
The values of $u, u_1, v, v_1, n, n_1, n_2$, etc., used in these two proofs are unrelated to one another.
Slightly more detailed answer:
In part (1) of the proof, they are proving $P \iff Q$ where $$\begin{align*}P &: \text{ For all } u,u_1,v,v_1 \in G, \quad \Big(u_1N=uN,\ \ v_1N=vN \implies (u_1v_1)N = (uv)N\Big)\\ Q &: \text{ For all } g\in G, n \in N, \quad \Big(gng^{-1} \in N\Big)\end{align*}$$
In the first half of the proof, they are proving $P \implies Q$, so they assume $P,$ let $g \in G, n \in N$ be arbitrary and set $u = 1, u_1 = n, v=v_1 = g^{-1}$. They then show $gng^{-1} \in N$ which proves $Q.$ This completes their proof of $P \implies Q.$
In the second half of the proof, starting with "Conversely, assume ...", they are no longer dealing with any of the variable assignments used in the first half. Rather, they assume $Q$ is true and let $u,u_1,v,v_1 \in G$ be arbitrary such that $u_1 \in uN, v_1 \in vN$. From here, they eventually prove that $(u_1v_1)N = (uv)N$. From here, we conclude that the implication in $P$ is true, and since $u,v,u_1,v_1 \in G$ were arbitrary in this part of the proof, we can conclude $P$ itself is true. This completes their proof of $Q \implies P.$
Having shown the two proofs of $P \implies Q$ and $Q \implies P,$ they conclude their proof of $P \iff Q.$
Because what to be proved is what in Proposition 5 itself which is $uN \: vN = uvN$, NOT $uvN = u_1 v_1 N$. The later is what a well defined condition is.
By letting $u = 1$, $u_1 = n$, and $v = v_1 = g^{-1}$ basically we first prove that $N \: vN = vN \Rightarrow gng^{-1} = n$, where $1, n \in N$ and $v = g^{-1}$, using that well defined condition to arrive at $gng^{-1} = n$ as asserted.
That result implies that:
$uN \: vN = u (N \: vN) = uvN$