It does, if you identify $T^*M \otimes TM$ with $\text{Hom}(TM, TM)$. I'll describe the identification in the linear algebra setting between $V^* \otimes W$ and $Hom(V,W)$. Pick a basis $(v_1, ..., v_n)$ for $V$, take $(\phi^1, ..., \phi^n)$ to be the dual basis of $(v_i)$ for $V^*$ and take $(w_1, ..., w_m)$ to be a basis of $W$. Elements of $V^* \otimes W$ are of the form $\sum a_i^j \phi^i \otimes w_j$. How can we interpret them as maps from $V$ to $W$? Given a vector $v \in V$, the action of this element on $v$ is $ \left( \sum a_i^j \phi^i \otimes w_j \right)\left( v \right) = \sum a^j_i \phi_i(v)w_j. $ If we plug in $v_i$ we see that $ \left( \sum a_i^j \phi^i \otimes w_j \right)\left( v_i \right) = \sum_{j} a^j_i w_j. $ That is, the map defined by $\sum a_i^j \phi^i \otimes w_j$ is just the map represented in the bases $v$ and $w$ by the matrix $(a^j_i)$.
In the other direction, given a map $T : V \rightarrow W$, represent it in the bases $v$ and $w$ by $a^j_i$ and match it to the element $\sum a_i^j \phi^i \otimes w_j$. Just like you expand a vector $w \in W$ in a basis $w_i$ as $w = \sum_j a^j w_j$, where the coefficients are scalars, you expand a map $T$ as $T = \sum a_i^j \phi^i \otimes w_j $ where the "coefficients" $\sum_{i} a_i^j \phi^i$ are functionals in $V^*$.
As usual, everything passes from the linear algebra setting to the vector bundle setting, and you identify a tensor $\sum a_i^j dx^i \otimes \frac{\partial}{\partial x^j}$ with a map between $TM$ and $TM$, given in the frame $(\frac{\partial}{\partial x^j})$ by the matrix $(a_i^j)$.