I'm not sure what you're looking for, but maybe this will help. I am completely avoiding any mention of coordinates or bases.
A connection on a vector bundle determines and is determined by the corresponding covariant derivative on sections of $V$.
Connections on vector bundles $V$ and $W$ determine a connection on $V\otimes W$. In terms of covariant derivatives the rule is $\nabla(v\otimes w)=\nabla v\otimes w+v\otimes \nabla w$, just like the usual product rule.
A connection on $V$ determines a connection on the dual $V^\ast$. If we write $<\omega,v>$ for the result of evaluating $\omega\in V^\ast$ on $v\in V$, then the rule is
$$ d<\omega,v>=<\nabla\omega,v>+<\omega, \nabla v>.$$ Here $d$ denotes the usual derivative of functions; if we like we can call it the covariant derivative for the trivial connection on the trivial bundle $1$ with fiber $\mathbb R$. This equation may then be read as saying that the canonical "contraction" map $V\otimes V^\ast\to 1$ is compatible with connections: the result is the same whether you first contract $\omega\otimes v$ and then differentiate or first differentiate and then contract. (Of course, the equation may also be used as a definition of the connection on $V^\ast$ determined by a connection on $V$, by rewriting: $<\nabla\omega,v>=d<\omega,v>-<\omega, \nabla v>$.)
Connections on $V$ and $W$ also determine a connection on $Hom(V,W)$. The rule can be given as $\nabla(Tv)=(\nabla T)(v)+T(\nabla v)$. This generalizes the rule for $V^\ast=Hom(V,1)$. It also agrees with the rule for tensor products, if we identify $Hom(V,W)$ with $V^\ast\otimes W$. Note that a section $T$ of $Hom(V,W)$ satisfies $\nabla T=0$ if and only if this map $T:V\to W$ satisfies $\nabla (Tv)=T(\nabla v)$.
The contraction operator $\omega\otimes v\mapsto <\omega,v>$ is a section of $Hom(V\otimes V^\ast,1)$. The displayed equation says that the covariant derivative of this section has zero.
In general a connection on $V$ determines connections on all the bundles you can make from $V$ by dualizing and tensoring (and symmetrizing). It works out that every contraction operator, for example the various maps $V\otimes V\otimes V\otimes V^\ast \otimes V^\ast\to V\otimes V\otimes V^\ast$, will have zero derivative,
The usual connection on a Riemannian manifold is chosen to be compatible with the inner product in the sense that $d(v.w)=\nabla v.w+v.\nabla w$. In other words, it is chosen so that the inner product, as a section of $T^\ast\otimes T^\ast=(T\otimes T)^\ast$, has zero derivative.