I have troubles following the book "Elements of Noncommutative Geometry". Let $E\to M$ be a vector bundle. Then a connection on $E$ is a linear map $\nabla:\Gamma^\infty(M,E)\to\Gamma^\infty(M,E)\otimes_{C^\infty(M)}\Omega^k(M)$ satisfying $$\nabla(s\otimes\omega)=(\nabla s)\otimes\omega + s\otimes d\omega\,$$ for $s\in\Gamma^\infty(M,E),\omega\in\Omega^\bullet(M)$. (Note $\Gamma^\infty$ are the smooth sections)
Consider the connection $\nabla^g$ on $TM$. Then we can also define a connection on the cotangent bundle. Since $\Omega^1(M)=\mathrm{Hom}_{C^\infty(M)}(\mathfrak{X}(M),C^\infty(M))$ holds, a function $f\in C^\infty(M)$ can be written as $f=\alpha(X)$ with $\alpha\in\Omega^1(M), X\in\mathfrak{X}(M)$. Applying the Leibniz rule, one obtains
$\nabla^g \alpha(X)= d\,\alpha(X)-\alpha(\nabla^gX)$
My idea was to write $\alpha(X)=X\otimes\alpha$ and then applying the Leibniz rule, but I obtain a plus and not a minus. How do I get the minus?
Thanks for your help.
Somehow things seem to be slightly messed up in your question. Initially, one defines a connection as sending $\Gamma^\infty(M,E)$ to $\Gamma^\infty(M,E)\otimes_{C^\infty(M,\mathbb R)}\Omega^1(M)$ such that $\nabla(fs)=f\nabla s+s\otimes df$. (Indeed, I would view it as easier to view the left hand side as $\Gamma^\infty(M,E\otimes T^*M)$, but that is a different story.) You can extend this to an operator $\Gamma^\infty(M,E)\otimes_{C^\infty(M,\mathbb R)}\Omega^k(M)\to \Gamma^\infty(M,E)\otimes_{C^\infty(M,\mathbb R)}\Omega^{k+1}(M)$, called the covariant exterior derivative, which indeed satisfied a Leibniz rule involving the exterior derivative on forms of higher degree. But I don't really think that this is really related to what you want to know.
The story about the cotangent bundle goes into a different direction: For any vector bundle $E\to M$ there is a dual bundle $E^*\to M$ and a linear connection on $E$ induces a linear connection on $E^*$. The construction of this connection is usually viewed in a broader context, namely that one can define connections on tensor products of bundles via yet another Leibniz rule which essentially says that $\nabla_X(s_1\otimes s_2)=(\nabla_X s_1)\otimes s_2+s_1\otimes\nabla_X s_2$. Now appling this to $\alpha\otimes Y$, one obtains $(\nabla_X\alpha)\otimes Y+\alpha\otimes\nabla_XY$. The contraction maps this tensor field to $ (\nabla_X\alpha)(Y)+\alpha(\nabla_XY)$. But contracting $\alpha\otimes Y$, one obtains the smooth function $\alpha(Y)$ and the only reasonable covariant derivative on smooth functions is the standard action of vector fields. Hence one obtains $X(\alpha(Y))= \nabla_X\alpha)(Y)+\alpha(\nabla_XY)$, which prescisely gives the usual formula for $\nabla_X\alpha$.