Existence of a vector field which dominates the first local vector fields given by the charts of a locally finite covering Let $M$ be a smooth manifold, let $\{U_i,\psi_i\}_{i\in I}$ be locally finite family of charts and let $K_i\subseteq U_i$ be compact subsets.
Does there exist a vector field $X$ on $M$, such that $$|\frac{\partial}{\partial\psi_i^1}(f)(x)|\le|X(f)(x)|$$ for all $i\in I$, $x\in K_i$ and $f\in\mathcal{C}^\infty(M,\mathbb{R})$, where $\frac{\partial}{\partial\psi_i^1}$ denotes the first local vector field of the chart $\psi_i$?
I tried to prove this with a partition of unity, but unfortunately was not very successful.
 A: Unless the coordinate vector fields $\partial/\partial \psi_i^1$ are all scalar multiples of each other where the charts overlap, the answer is no. Here's a counterexample. Let $M=\mathbb R^2$, $U_1 = U_2 =\mathbb R^2$, $\psi_1(x,y) = (x,y)$, and $\psi_2(x,y) = (y,x)$.  Then the vector fields $\partial/\partial \psi_i^1$, $\partial/\partial \psi_i^2$ span the tangent space at each point of $\mathbb R^2$. No matter what the vector field $X$ is, there will be some connected open set $V$ on which you can find a nonconstant smooth function $f$ such that $X(f)\equiv 0$ (and then, if you wish, extend this arbtrarily to a global smooth function on $\mathbb R^2$).  If your inequality held on $V$, it would imply that $df\equiv 0$ on $V$, and hence $f$ would be constant there, a contradiction.
EDIT: The OP asked in a comment if there always exists a locally finite cover by charts $(U_i,\psi_i)$ with compact subsets $K_i$ such that the interiors of $K_i$ still cover $M$, such that a vector field as described in the original question does exist for this specific chosen family of charts.
The answer is yes if and only if $M$ admits a nonvanishing vector field.
If not, then every vector field $X$ on $M$ must have a zero at some point, and no matter what charts you choose, at that point the inequality must fail for some $f$. 
On the other hand, suppose there is a nonvanishing vector field $X$ on $M$. The canonical form theorem for nonvanishing vector fields (sometimes called the "straightening lemma") shows that in a neighborhood of each point, there exists a coordinate chart $(U,\psi)$ such that $X=\partial/\partial \psi^1$ there. By shrinking $U$ if necessary, we can arrange for it to have compact closure.  Let $(U_i,\psi_i)$ be a locally finite refinement of this cover, with the charts $\psi_i$ defined by restricting the original charts.  Then, because $M$ is normal, for each $i$ there exists an open subset $V_i$ such that $\overline V_i\subset U_i$ and $V_i$ contains the complement of the union of all $U_j$'s for $j\ne i$. The sets $K_i = \overline V_i$ do the trick.
