= Solution
First establish the scalar <Leibniz rule>, since it is needed for an extension to all <tensor fields>. Suppose every component of $M$ has positive dimension. Computing $D(fgY)$ first with scalar $fg$ and then as $f(gY)$ gives
$$
\bigl(D(fg)-fDg-gDf\bigr)Y=0.
$$
At any point, a <smooth cutoff function> times a coordinate <vector field> can be chosen nonzero there. Evaluating the displayed identity proves
$$
D(fg)=fDg+gDf,\qquad D1=0.
$$
Thus the scalar operator is a <derivation of an algebra>. <Derivations of smooth functions are vector fields>, so the scalar operator is differentiation along a unique <vector field> $Z$; no continuity assumption is needed. Indeed, if $h$ vanishes near $p$, take a <smooth cutoff function> $\chi$ equal to one near $p$ and supported where $h=0$. The identity $D(\chi h)=(D\chi)h+\chi Dh=0$ gives $Dh(p)=0$. Thus $D$ depends only on the local function germ, so local coordinate functions may be extended with cutoffs before applying it. The local identity $h(x)=h(p)+\sum_i(x^i-x^i(p))h_i(x)$ gives $Dh(p)=\sum_i D(x^i)(p)\partial_i h(p)$. Smooth local coefficients $D(x^i)$ define $Z$.
The vector-field operator is local as well. If $Y$ vanishes near $p$, choose a cutoff $h$ equal to one near $p$ with support where $Y=0$. Then $D(hY)=hDY+(Dh)Y$ gives $DY(p)=0$. We can therefore work with local frames without presuming a global frame.
For a <differential one-form> $\omega$, the only possible contraction-compatible definition is
$$
\boxed{(D\omega)(Y)=D(\omega(Y))-\omega(DY).}
$$
The scalar product rule and the given vector-field rule show that this is linear over $C^\infty(M)$ in $Y$, so it defines a one-form. Its coefficients are smooth by evaluating the formula on local frame fields extended with cutoffs. It is real-linear in $\omega$, and direct substitution gives $D(f\omega)=fD\omega+(Df)\omega$.
The original PDF's hint has a transpose error. If $De_i=\sum_j a_{ij}e_j$ for $e_i=\partial/\partial x^i$, then evaluation on every $e_j$ forces
$$
\boxed{D(dx^i)=-\sum_j a_{ji}\,dx^j,}
$$
not the untransposed coefficient array printed in the hint. For example, take $Df=0$, $De_1=e_2$ and $De_2=0$ on $\mathbb R^2$. The correct values are $D(dx^1)=0$, $D(dx^2)=-dx^1$. The printed hint instead makes the derivative of $dx^1(e_2)=0$ equal to $-1$.
There is also a genuine zero-dimensional edge case in the hypotheses: on a one-point manifold the vector-field space is zero, so the printed rule imposes no condition on the scalar map. Taking $Df=f$ satisfies that rule but cannot extend to a <tensor derivation>, since the product rule requires $D1=0$. Thus the extension theorem is valid on positive-dimensional manifolds as above, or in every dimension if the scalar product rule is added explicitly. The next two parts use these precise hypotheses.
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