Let $\rho$ be a non-negative smooth scalar density on $M$, and $\mu$ the associated measure. Recall that this means (almost by definition) that given any chart $(U,\alpha)$ on $M$ and any Lebesgue-measurable subset $A\subset M$, we have
\begin{align}
\mu(A\cap U)&=\int_{\alpha[A\cap U]}\rho_{\alpha}(x)\,d\lambda_n(x)\tag{$*$},
\end{align}
where $\rho_{\alpha}:\alpha[U]\to (0,\infty)$ is the local coordinate expression of the density and $\lambda_n$ denotes the Lebesgue measure on $\Bbb{R}^n$. More succinctly, this says that on $\alpha[U]$, we have $\frac{d(\alpha_*\mu)}{d\lambda_n}=\rho_{\alpha}$, or equivalently that when restricted to $U$, we have $\frac{d\mu}{d(\alpha^*\lambda_n)}= \rho_{\alpha}\circ \alpha$ (where $\alpha_*$ denotes pushforward of measures, and $\alpha^*=(\alpha^{-1})_*$ is the pullback of measures).
With this in mind, the first definition of null implies the second. To justify this, fix an at-most countable atlas $\{(U_i,\alpha_i)\}$ of $M$, and suppose $A$ is a null set in the sense of the first definition. Then for each index $i$, we have $\mu(A\cap U_i)=\int_{\alpha_i[A\cap U_i]}\rho_{\alpha_i}\,d\lambda_n=0$, since we're integrating a non-negative function over a set of Lebesgue measure zero. Therefore (countable subadditivity of measures), $\mu(A)\leq \sum_{i}\mu(A\cap U_i)=0$.
The converse is true if we assume $\rho$ is a strictly positive density. In this case, if $\mu(A)=0$, then for any chart $(U,\alpha)$ of $M$, we have by monotonicity of measures, $0=\mu(A\cap U)=\int_{\alpha[A\cap U]}\rho_{\alpha}\,d\lambda_n$; and since $\rho_{\alpha}$ is a strictly positive function, we have $\lambda_n(\alpha[A\cap U])=0$, thereby recovering the first definition.
The analogous statements for full measure can be obtained from the above statements by taking complements (after all a set is defined to have full measure if its complement has measure zero).
For the final question, we have the following theorem:
Let $M$ be a smooth manifold, $\rho_1,\rho_2$ smooth non-negative densities on $M$, and $\mu_1,\mu_2$ their associated measures. If $\rho_2$ is strictly positive, then $\mu_1\ll\mu_2$, and the Radon-Nikodym derivative $\frac{d\mu_1}{d\mu_2}:M\to \Bbb{R}$ is a smooth non-negative function such that for any coordinate chart $(U,\alpha)$, we have
\begin{align}
\frac{d\mu_1}{d\mu_2}\bigg|_U &=\left(\frac{\rho_{1,\alpha}}{\rho_{2,\alpha}}\right)\circ \alpha.
\end{align}
The proof is pretty simple. If $\mu_2(A)=0$, then because $\rho_2$ is assumed strictly positive, we have that $A$ is a null set in the sense of the first definition, and hence $\mu_1(A)=0$ (because $\rho_1$ is a non-negative density). This shows $\mu_1\ll\mu_2$, and hence, by the Radon-Nikodym theorem, the derivative $\frac{d\mu_1}{d\mu_2}$ exists.
However, in this specific case, we do not have to appeal to the Radon-Nikodym theorem to assert the existence of the derivative; we can actually proceed directly. If we fix a chart $(U,\alpha)$, then the definition $(*)$ above shows tells us that $\frac{d\mu_1}{d(\alpha^*\lambda_n)}=\rho_{1,\alpha}\circ \alpha$ and $\frac{d\mu_2}{d(\alpha^*\lambda_n)}=\rho_{2,\alpha}\circ \alpha$. Now hopefully you've justified the 'chain rule' for Radon-Nikodym-derivatives, because from this it follows $\frac{d\mu_1}{d\mu_2}\bigg|_U=\frac{\rho_{1,\alpha}}{\rho_{2,\alpha}}\circ \alpha$ (and from this explicit formula, we know the Radon-Nikodym derivative is smooth and non-negative).
Finally, if you assume the two densities $\rho_1,\rho_2$ are strictly positive (which is the case if the densities are induced by a non-degenerate $(0,2)$-tensor field on the manifold, e.g. from a (pseudo-)Riemannian metric, or a symplectic form), then applying the theorem above twice, we see that the induced measures are mutually absolutely continuous, and the Radon-Nikodym derivatives are as obvious as it can get.
