The physicist's way: use standard formulas
(BEST!)
If $x=x(y)$ compute:
$$\sum_i
dx^i\otimes\frac{\partial}{\partial x^i}=\sum_i
(\sum_j
\frac{\partial x_i}{\partial y^j} dy^j
\otimes \sum_k\frac{\partial y^k}{\partial x^i} \frac{\partial}{\partial y^k})=\sum_i (\sum_{j,k}\frac{\partial x_i}{\partial y^j}\frac{\partial y^k}{\partial x^i}dy^j\otimes \frac{\partial}{\partial y^k})\\=\sum_{j,k} \delta ^k_jdy^j\otimes \frac{\partial}{\partial y^k}=\sum_j dy^j\otimes \frac{\partial}{\partial y^j}$$ and notice that you end up with the same expression in the $y$ coordinates.
The pure mathematician's way (VERY CHIC!)
We have a canonical isomorphism of vector bundles on $M$: $$HOM(T_M,T_M) \stackrel \sim \to T^*_M\otimes T_M $$ Taking global sections we obtain an isomorphism of $ C^\infty (M)$-modules $$\varphi:\Gamma(M, HOM(T_M,T_M))=Hom(T_M,T_M)\stackrel \sim \to \Gamma(M, T^*_M\otimes T_M)$$
The identity morphism on the left $[Id:T_M\stackrel = \to T_M]\in Hom(T_M,T_M)$ is sent to the tensor field $\varphi (Id)=C\in \Gamma(M, T^*_M\otimes T_M)$ on the right.
And this $C$ is your mysterious tensor field: its expression in a local chart $(U,x)$ with domain $U\subset M$ is $$C\vert U=\sum_i
dx^i\otimes\frac{\partial}{\partial x^i}\in \Gamma(U, T^*_M\otimes T_M)$$
A personal point of view
As an algebraic geometer used to the abstract tools of scheme theory and as a Bourbaki fan, my natural tendency is to prefer the second point of view.
However in order to keep myself honest I try to occasionally compute the way the physicists (for example the general relativists) do.
Of course Spivak's fantastic treatise convinced me once and for all that one can give a completely rigorous explanation for the sort of computation in the first answer, so that now I do these computations purely algorithmatically, with my brain completely disengaged...
