The definition of the tautological 1-form is the following:
Let $Q$ be a smooth manifold, the tautological 1-form is a 1- form on $T^{*}Q$, so a smooth map $M \rightarrow T^{*} Q$.
The definition is the following: A point $x \in T^{*}Q$ has projection $q= \pi(x) \in Q$ represents a linear 1-form \begin{align*} \alpha: T_qQ \rightarrow \mathbb{R} \end{align*} Now $\lambda_{can}$ is defined as \begin{align*} (\lambda_{can})_x(v) = \alpha(\pi_{*} v) \ \ \ \ \ \ (*) \end{align*} where $\pi : T^{*}Q \rightarrow Q $ is the bundle projection and $\pi_{*}: T(T^{*}Q) \rightarrow TQ$ its differential.
Now what I don't understand: On the left hand side of (*), $v$ should be in $T_xQ$ but on the right hand side, it should be in $T(T^{*}Q)$.
Can somebody explain this to me?
A $1$-form $\lambda$ on $T^*Q$ is a section of the cotangent bundle of $T^*Q$, so a map $\lambda:T^*Q\to T^*(T^*Q)$. So $\lambda$ has first to take a point $x\in T^*Q$ (with $\pi(x)=q$, which means that as a function you have $x:T_qQ\to\mathbb{R}$), then a tangent vector $v\in T_x(T^*Q)$. Thus
$${\lambda_{can}}_x(v)=x(\pi_*v)$$
makes sense since $\pi_{*,x}v$ is in $T_qQ$.