I am currently studying Riemannian geometry via doCarmo's book, in which he introduces the affine connection $$\nabla: \Pi(M) \times \Pi(M) \to \Pi(M), \text{ where } \Pi(M) \text{ is the space of vector fields on M.}$$
From this we can introduce the covariant derivative $$\frac{D}{dt}: \Pi(M) \to \Pi(M). $$
I have encountered a more general definition of a connection for vector bundles (Vector Bundles with a Connection, S.S Chern) where the connection is given by $$ D:\Gamma(E)\rightarrow\Gamma(T^*M\otimes{E}), \text{ where } \Gamma(*) \text{ denotes the vector space of sections of } *, $$ and am trying to reconcile the two notions. In particular my confusion lies in the question of why the image $\Gamma(T^*M\otimes{E})$ is as it is, and I would be grateful for any clarifications made on this.
An element $\xi \in \Gamma(T^*M \otimes E)$ can be interpreted as a map $\Gamma(TM) \to \Gamma(E)$ in a natural way - given any vector field $X$, just contract it with the $T^*M$ slot of $\xi$: if $\xi = \xi^\alpha_i dx^i \otimes E_\alpha$ then $$\xi(X) = \xi^\alpha_i X^i E_\alpha.$$
A brief digression: In fact $\Gamma(T^*M \otimes E)$ is exactly the set of maps $\Gamma(TM) \to \Gamma(E)$ that are pointwise linear; i.e. for each $p$, $\xi(X)_p$ is a linear function of $X_p$. The intuition here is that $\xi$ should be the differential of a section of $E$, and plugging in a vector tells you the corresponding directional derivative; so this pointwise linearity is a reasonable requirement.
Back to your question: We can now view $D : \Gamma(E) \to \Gamma(T^*M \otimes E)$ as a map $\Gamma(E) \times \Gamma(TM) \to \Gamma(E)$ by "uncurrying": simply define $D(s,X) = D(s)(X)$, where the action of $D(s)$ on $X$ is as I described above. The two arguments here are now the section to be differentiated and the direction to differentiate it in.
In the case where $E=TM$ this becomes $\Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)$ as you wanted.