0th-order differential operator vs. 1st-order differential operator on a vector bundle $(E, \pi,M)$ Consider a vector bundle $(E,\pi,M)$. A 0th-order differential operator on $E$ is a $C^\infty(M)$-linear endomorphism $\Gamma E\rightarrow\Gamma E$. $\Gamma E$ is the set of sections on $M$. 
A 1st-order differential operator on $E$ is an $\mathbb R$ linear map $D:\Gamma E\rightarrow \Gamma E$ such that for each $u\in C^\infty(M)$, the map
$\Gamma E\rightarrow\Gamma E,\quad \mu\mapsto D(u\mu)-uD(\mu)$
is a 0th-order differential operator. 
But it seems to me it is impossible to construct the operator $D$ from a 0th-order differential operator. Consider a smooth function $f\in C^\infty(M)$, then $f\mu\in\Gamma E$, therefore,
$f\mu\mapsto D(uf\mu)-uD(f\mu), \forall u\in C^\infty(M)$     ------- Eq.(1)
But $uf$ is an arbitrary smooth function on $M$, 
$f\mu\mapsto D(uf\mu)-ufD(\mu)$    ------- Eq.(2)
Comparing the two equations Eq.(1) and (2) gives 
$uD(f\mu)=ufD(\mu)\Rightarrow D(f\mu)=fD(\mu)$
However, $f$ is actually arbitrary, so $\mu\mapsto D(u\mu)-uD(\mu)=0, \forall u\in C^\infty(M)$.
Probably I misunderstand the definitions? Please help me out. Thank you!

Another question, how to prove the map $\mu\mapsto D(u\mu)-uD(\mu)$ is $C^\infty(M)$ linear?
 A: I couldn't follow your derivation that $D(u\mu) - uD(\mu)=0$.  But I think you're misinterpreting the definition. It means that $D$ is a first-order differential operator if and only if for each $u\in C^\infty(M)$, the map $F_u\colon \Gamma E \to \Gamma E$ defined by $F_u(\mu) = D(u\mu) - u D(\mu)$ is linear over $C^\infty(M)$ in $\mu$. More specifically, this means that the following identity holds for all $u,f\in C^\infty(M)$ and $\mu\in \Gamma E$:
$$
D(uf\mu) - u D(f\mu) = f\big(D(u\mu)) - u D(\mu)\big).
$$
As for your second question, how you would prove it depends on specifically what the operator $D$ is.  Here's an example.  Let $E = TM$, the tangent bundle of $M$, and let $X$ be a smooth vector field on $M$. Consider the map $\mathscr L_X\colon \Gamma(TM)\to \Gamma(TM)$, i.e., Lie differentiation by $X$.  A straightforward computation based on the properties of the Lie dervative shows that for all $u,f\in C^\infty(M)$ and all $Y\in \Gamma(TM)$, 
$$
\mathscr L_X(ufY) - u \mathscr L_X(fY) = f\big( \mathscr L_X(uY) - u \mathscr L_XY\big).
$$
Thus $\mathscr L_X$ is a first-order differential operator.
