Partition of unity and volume form on a manifold

A convex combination of volume forms could turn out to be zero, violating the nowhere-vanishing condition. For example, $$\frac{1}{2}dx \wedge dy + \frac{1}{2}(-dx) \wedge dy = 0$$

This is actually exactly what will happen if you try to do the proposed procedure on a mobius strip with standard parameterization.

In contrast, at each point a metric corresponds to a positive definite matrix (or operator more generally), and convex combinations of positive definite matrices are themselves positive definite.

Suppose $p \in U_1\cap U_2$ and consider $\rho_1(p)\phi_1^*(\operatorname{vol})_p + \rho_2(p)\phi_2^*(\operatorname{vol})_p \in \bigwedge^nT_p^*M$. While $\phi_1^*(\operatorname{vol})_p$ and $\phi_2^*(\operatorname{vol})_p$ are non-zero elements of $\bigwedge^nT_p^*M$, they could be negative multiples of each other, in which case $\rho_1(p)\phi_1^*(\operatorname{vol})_p + \rho_2(p)\phi_2^*(\operatorname{vol})_p$ could be zero.

For example, consider the manifold $S^1$. Let $U_1 = S^1\setminus\{1\}$, $\phi_1 : U_1 \to (0, 2\pi)$, $e^{i\theta}\mapsto \theta \bmod{2\pi}$, and $U_2 = S^1\setminus\{-1\}$, $\phi_2 : U_2 \to (-\pi, \pi)$, $e^{i\theta} \mapsto \pi - \theta \bmod{2\pi}$. Then $\phi_1^*(dx) = d\theta$ and $\phi_2^*(dx) = d(\pi - \theta) = -d\theta$ so $\rho_1(p)\phi_1^*(dx)_p + \rho_2(p)\phi_2^*(dx)_p = (\rho_1(p) - \rho_2(p))d\theta$ which is zero whenever $\rho_1(p) = \rho_2(p) = \frac{1}{2}$.