how does $PATH$ being $NL$-complete help to prove $NL = coNL$
$\mathsf{NL}$ consists of all languages accepted by nondeterministic Turing machines using logarithmic space. One way to think of such machines is as follows:


The input is written on a read-only input tape.
There are one or more work tapes of logarithmic size.
There is a read-only oracle tape on which the head is constrained to only move to the right.


We can think of each such machine as computing a Boolean function on two arguments: $x$, which is the actual input, and $y$, which is the initial contents of the oracle tape. Such a machine $M$ nondeterministically computes the language $L(M)$ given by:
$$
x \in L(M) \Longleftrightarrow \exists y \text{ s.t. $M(x,y)$ accepts}.
$$
In other words, a language $L$ is in $\mathsf{NL}$ if there exists a logspace Turing machine $M$ with an auxiliary oracle input such that
$$
x \in L \Longleftrightarrow \exists y \text{ s.t. $M(x,y)$ accepts}.
$$

A language is in $\mathsf{coNL}$ if its complement is in $\mathsf{NL}$. Equivalently, a language $L$ is in $\mathsf{coNL}$ if there exists a logspace Turing machine $M$ with an auxiliary oracle input such that
$$
x \notin L \Longleftrightarrow \exists y \text{ s.t. $M(x,y)$ rejects}.
$$
Equivalently,
$$
x \in L \Longleftrightarrow \forall y \text{ $M(x,y)$ accepts}.
$$

Suppose now that $L$ is in $\mathsf{NL}$, say $L = L(M)$, and let $M'$ be obtained from $M$ by complementing the output. Then
$$
x \in L(M') \Longleftrightarrow \exists y \text{ s.t. $M'(x,y)$ accepts}
\Longleftrightarrow \exists y \text{ s.t. $M(x,y)$ rejects}.
$$
As you can see, $L(M')$ isn't necessarily the complement of $L(M)$. Indeed, given any machine $M$, we can come up with another machine $\tilde{M}$ such that $L(M) = L(\tilde{M})$, but $\tilde{M}(x,y_0)$ always rejects, for some fixed $y_0$ (for example, $\tilde{M}(x,0\Sigma^*)$ always rejects, and $\tilde{M}(x,1y) = M(x,y)$). In this case, $L(\tilde{M}') = \Sigma^*$.



Now I can answer your questions:


$\overline{\mathsf{PATH}} \in \mathsf{coNL}$ since $\mathsf{PATH} \in \mathsf{NL}$ and by definition $\mathsf{coNL}$ consists of the complements of all languages in $\mathsf{NL}$.
Furthermore, $\overline{\mathsf{PATH}}$ is $\mathsf{coNL}$-hard because $\mathsf{PATH}$ is $\mathsf{NL}$-hard. Indeed, suppose that $L \in \mathsf{coNL}$. Then $\overline{L} \in \mathsf{NL}$. Since $\mathsf{PATH}$ is $\mathsf{NL}$-hard, there is a logspace reduction $f$ such that $x \in \overline{L}$ iff $f(x) \in \mathsf{PATH}$. Thus $x \in L$ iff $f(x) \in \overline{\mathsf{PATH}}$, and so $f$ also reduces $L$ to $\overline{\mathsf{PATH}}$.
Flipping the output value of a nondeterministic machine does not result in complementing the accepted language.


Elaborating on the second point, for some nondeterministic complexity classes $\mathsf{X}$, it is known that $\mathsf{X} \neq \mathsf{coX}$. The simplest example is $\mathsf{RE} \neq \mathsf{coRE}$: the halting problem is in $\mathsf{RE}$ but not in $\mathsf{coRE}$. Therefore $\mathsf{NL} = \mathsf{coNL}$ is a nontrivial statement.