Since no one has written an official answer to this question and the bounty is ending in a few hours, I'll provide an answer of my own to the question, though see ulrich's comments for another approach.

1) The crucial observation here is that $\mathfrak m_x=\pi_*(\mathcal O_{C'}(-x_1-x_2))$ where $x\in C$ is a node, $\pi:C'\rightarrow C$ is the partial normalization at $x$ and $x_1,x_2$ are the points in $C'$ lying over $x$.  This is fairly clear from the description of $\mathcal O_C(V)$ as the set of $ f\in \mathcal O_{C'}(\pi^{-1}(V))$ such that $f(x_1)=f(x_2)$ for any affine $V$ such that $x$ is its only singular point.  Then the claim is seen as follows:

Writing $\mathcal L=(\mathcal L\otimes \omega_C^{-1})\otimes \omega_C$, we get that $$\text{Hom}(\mathfrak m_x,\mathcal L)\cong \text{Hom}(\mathfrak m_x (\mathcal L^{-1}\otimes \omega_C),\omega_C)\cong H^1(C,\mathfrak m_x(\mathcal L^{-1}\otimes \omega_C))^{\vee}.$$  Then using the crucial observation above, the projection formula, and the fact that $\pi$ is finite so the higher direct images vanish, we see that 

$H^1(C,\mathfrak m_x (\mathcal L^{-1}\otimes \omega_C))^{\vee}\cong H^1(C',\mathcal O_{C'}(-x_1-x_2)\otimes\pi^*\mathcal L^{-1}\otimes \pi^*\omega_C)\cong H^0(C',\pi^*\mathcal L),$

where I've used the fact that $\pi^*\omega_C\cong \omega_{C'}(x_1+x_2)$ and Serre duality on $C'$.  The proof of the $\mathfrak m_x^2$ case is almost identical.

2) They choose $n\geq 2$ since when they used Serre duality an exponent of $1-n$ appeared for $\omega_C$, and $n$ was originally $n\geq 3$, so they WLOG assume now $n$ is the exponent of $\omega_C$ but $n\geq 2$.  For the rest, a possibly clearer proof of this result is given as Lemma 10.6.1 in "Geometry of Algebraic Curves, Vol. II" by ACG.

3) The projective bundle is not necessarily trivial, and the Hilbert scheme constructed there parametrizes only those families for which a trivialization exists.  However, any family can be pulled back to the principal $PGL$ bundle associated to the projective bundle, and there it does become trivial.  Thus there is a map from the associated $PGL$-bundle to the Hilbert scheme.