There's actually a proof in Geometry of Algebraic Curves - a bit after formula 2.19. However this description is indirect and in particular doesn't use the normalization map. Here's a less elementary proof that does use it:

Let $X$ be a nodal curve, say with a single node $q$. Let $\nu :\tilde{X} \to X$ be its normalization with $\nu^{-1}(p)=\{p_1,p_2\}$. Let $\imath : \{q\} \hookrightarrow X$ and $\jmath : X \backslash q \hookrightarrow X$. Similarly $\tilde{\imath} : \{p_1,p_2\} \hookrightarrow \tilde{X}$ and $\tilde{\jmath} : \tilde{X} \backslash \{p_1,p_2\} \hookrightarrow \tilde{X}$. We get an exact triangle

$$\tilde{\imath}_*\tilde{\imath}^!\omega_{\tilde{X}} \to \omega_{\tilde{X}} \to \tilde{\jmath}_*\tilde{\jmath}^!\omega_{\tilde{X}}$$ or
$$\tilde{\imath}_*\omega_{\{p_1,p_2\}} \to \omega_{\tilde{X}} \to \tilde{\jmath}_*\omega_{\tilde{X} \backslash \{p_1,p_2\}}$$
and similarly for $X$. $\nu$ induces a map between the sequences, which is an isomorphism on the last term and a sum for the first one. This, together with the fact that the first term is concentrated in degree $0$ and the last two in degree $1$ implies that $\omega_X$ is the kernel of $\nu_* \tilde{\jmath}_*\omega_{\tilde{X} \backslash \{p_1,p_2\}} \xrightarrow{\text{res}} \nu_*\tilde{\imath}_*\omega_{\{p_1,p_2\}}[1] \xrightarrow{\text{sum}}\imath_*\omega_{q}[1]$
i.e. exactly elements of $\omega_{\tilde{X}}(\infty\{p_1,p_2\})$ with residue summing to zero and any higher order poles vanishing.