= Solution
Write a local frame as $e_a$, its <dual basis> as $\epsilon^a$, and set $De_a=A^b{}_a e_b$. The one-form definition gives $D\epsilon^a=-A^a{}_b\epsilon^b$. Repeated indices are summed. Every <tensor field> has a unique local expansion
$$
T=T_{a_1\ldots a_k}^{b_1\ldots b_l}\,\epsilon^{a_1}\otimes\cdots\otimes\epsilon^{a_k}\otimes e_{b_1}\otimes\cdots\otimes e_{b_l}.
$$
Define $DT$ by applying the scalar operator to its coefficient and applying $D$ to one frame factor at a time, adding all these terms. In components this is
$$
\boxed{(DT)_{a_1\ldots a_k}^{b_1\ldots b_l}
=Z\!\left(T_{a_1\ldots a_k}^{b_1\ldots b_l}\right)
-\sum_{r=1}^k A^c{}_{a_r}T_{a_1\ldots c\ldots a_k}^{b_1\ldots b_l}
+\sum_{s=1}^l A^{b_s}{}_cT_{a_1\ldots a_k}^{b_1\ldots c\ldots b_l}.}
$$
In each term only the indicated slot is replaced. This formula is real-linear and satisfies the <Leibniz rule> for a <tensor product>: differentiating a coefficient product uses the scalar product rule, and the list of differentiated frame factors splits into the two factors' lists.
To check that it is intrinsic, write $e$ as the row of frame fields and $\epsilon$ as the column of dual fields. Change frame by $e'=eP$, where $P$ is an invertible smooth <matrix>. Then
$$
A'=P^{-1}(AP+ZP),\qquad \epsilon'=P^{-1}\epsilon.
$$
Differentiating the inverse <matrix> gives $Z(P^{-1})=-P^{-1}(ZP)P^{-1}$; it cancels the extra frame-change terms in the dual factors. The same cancellation in each tensor slot makes the two local formulas agree. Thus they glue to a global <tensor derivation>.
In particular, for arbitrary tensor types, not necessarily the same type,
$$
\boxed{D(S\otimes T)=DS\otimes T+S\otimes DT.}
$$
The resulting type is the sum of their two covariant counts and their two contravariant counts. This also supplies the required real-linearity for every type, including functions as type $(0,0)$.
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