You differentiate in the direction of some vector (field). Feeding in the vector field is going to remove the $\otimes T^*M$ :)
**EDIT** : To be concrete, given the map $D\colon\Gamma(E)\to\Gamma(E\otimes T^*M)$, a section $s$, and a vector field $X$ on $M$, we have $Ds \in\Gamma(E\otimes T^*M)$ with $(Ds)(X) = \
abla_X s$. This is the "classic" covariant derivative in the direction of $X$. In coordinates, if $s_1,\dots, s_r$ give a (local) basis for sections of $E$ and $\omega^1,\dots,\omega^n$ give a (local) basis for sections of $T^*M$, $D$ is defined by $$D(s_i) = \sum_{\substack{1\le j\le r\\\ 1\le k\le n}} a_{ik}^j s_j\otimes\omega^k \quad\text{for some functions } a_{ik}^j.$$ Then the covariant derivative viewpoint is given by $$\
abla_X(s_i) = \sum_{j,k} a_{ik}^j \omega^k(X)s_j.$$