Fix a metric $g$ on a smooth, closed manifold $\mathcal{M}$. Take a finite subcover of the manifold from its atlas. Is it true that any smooth partition of unity subordinate to this cover has uniformly bounded derivatives in $L^p (\mathcal{M})$ for each $p \geq 1$ ? 



Partitions of Unity
A counterexample: Let $M$ be the unit circle. Let the two charts be the arcs $A = (0,2\pi)$ and $B = (\pi/2, 5\pi/2)$. For each $n$, consider the partition of unity subordinate to $\{A,B\}$ given by 


$$ \psi\_{A,n} = \sin^2 ( (2n+1) \theta ) $$


and


$$ \psi\_{B,n} = \cos^2 ( (2n+1) \theta ) $$


Check that $\psi\_{A,n} (0) = 0 = \psi\_{B,n} (\pi/2)$, and clearly $\psi\_{A,n} + \psi\_{B,n} = 1$. 


By the scaling property it is easy to check that 


$$ \| \partial^k\psi\_{A,n} \| = (2n+1)^k \| \partial^k\psi\_{A,1} \| $$


And using that 


$$ \sin^2(\theta) = \frac12 (1-\cos 2\theta) $$


you see immediately that your desired uniform bound is impossible. 


