[proofplan]
We prove the identity by evaluating both one-forms on an arbitrary tangent vector to $T^*U$ in the induced cotangent coordinate basis. The bundle projection $\tau:T^*U\to U$ is coordinatewise the projection $(q,p)\mapsto q$, so its differential keeps the base components and discards the fibre components. The defining formula for the tautological one-form then gives $\lambda_\beta(v)=\sum_i p_i a_i$, and the coordinate one-form $\sum_i p_i\,dq_i$ has exactly the same value on the same vector.
[/proofplan]
[step:Write the point and tangent vector in induced cotangent coordinates]
Fix a point $\beta\in T^*U$, and set $q:=\tau(\beta)\in U$. By definition of the induced cotangent coordinates, there are unique [real numbers](/page/Real%20Numbers) $p_1,\dots,p_n$ such that
\begin{align*} \beta=\sum_{i=1}^n p_i(dq_i)_q. \end{align*}
Let $v\in T_\beta(T^*U)$ be arbitrary. In the coordinate basis associated to $(q_1,\dots,q_n,p_1,\dots,p_n)$, there are unique real numbers $a_1,\dots,a_n,b_1,\dots,b_n$ such that
\begin{align*} v=\sum_{i=1}^n a_i\left(\frac{\partial}{\partial q_i}\right)_\beta+\sum_{i=1}^n b_i\left(\frac{\partial}{\partial p_i}\right)_\beta. \end{align*}
[/step]
[step:Compute the differential of the bundle projection]
In the induced coordinates, the projection $\tau:T^*U\to U$ is represented by
\begin{align*}
(q_1,\dots,q_n,p_1,\dots,p_n)\mapsto(q_1,\dots,q_n).
\end{align*}
Therefore its differential at $\beta$ satisfies
\begin{align*}
d\tau_\beta\left(\left(\frac{\partial}{\partial q_i}\right)_\beta\right)=\left(\frac{\partial}{\partial q_i}\right)_q
\end{align*}
and
\begin{align*}
d\tau_\beta\left(\left(\frac{\partial}{\partial p_i}\right)_\beta\right)=0
\end{align*}
for each $i\in\{1,\dots,n\}$. By linearity of $d\tau_\beta$,
\begin{align*}
d\tau_\beta(v)=\sum_{i=1}^n a_i\left(\frac{\partial}{\partial q_i}\right)_q.
\end{align*}
[guided]
The map $\tau:T^*U\to U$ sends a covector to its base point. In the coordinates on $T^*U$, a point is described by its base coordinates and fibre coordinates:
\begin{align*}
(q_1,\dots,q_n,p_1,\dots,p_n).
\end{align*}
The projection forgets the fibre coordinates and keeps the base coordinates:
\begin{align*}
(q_1,\dots,q_n,p_1,\dots,p_n)\mapsto(q_1,\dots,q_n).
\end{align*}
Thus the coordinate vector $\left(\frac{\partial}{\partial q_i}\right)_\beta$ changes the $i$th base coordinate, so its image under $d\tau_\beta$ is $\left(\frac{\partial}{\partial q_i}\right)_q$. The coordinate vector $\left(\frac{\partial}{\partial p_i}\right)_\beta$ changes only the fibre coordinate $p_i$, so its image under $d\tau_\beta$ is $0$. Applying this to
\begin{align*}
v=\sum_{i=1}^n a_i\left(\frac{\partial}{\partial q_i}\right)_\beta+\sum_{i=1}^n b_i\left(\frac{\partial}{\partial p_i}\right)_\beta
\end{align*}
and using linearity gives
\begin{align*}
d\tau_\beta(v)=\sum_{i=1}^n a_i\left(\frac{\partial}{\partial q_i}\right)_q.
\end{align*}
This is the key geometric point: the tautological form evaluates the covector $\beta$ only on the base-point component of the tangent vector to the cotangent bundle.
[/guided]
[/step]
[step:Evaluate the tautological one-form on the tangent vector]
Using the defining formula for the tautological one-form,
\begin{align*} \lambda_\beta(v)=\beta(d\tau_\beta(v)). \end{align*}
Substituting the coordinate expression for $d\tau_\beta(v)$ gives
\begin{align*} \lambda_\beta(v)=\beta\left(\sum_{i=1}^n a_i\left(\frac{\partial}{\partial q_i}\right)_q\right). \end{align*}
Since $\beta=\sum_{j=1}^n p_j(dq_j)_q$ and $(dq_j)_q\left(\left(\frac{\partial}{\partial q_i}\right)_q\right)=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta, we obtain
\begin{align*} \lambda_\beta(v)=\sum_{i=1}^n p_i a_i. \end{align*}
[/step]
[step:Compare with the coordinate one-form]
Let $\alpha\in\Omega^1(T^*U)$ denote the coordinate one-form
\begin{align*} \alpha=\sum_{i=1}^n p_i\,dq_i, \end{align*}
where $q_i$ denotes the pulled-back coordinate function $q_i\circ\tau:T^*U\to\mathbb R$. Since $dq_i\left(\left(\frac{\partial}{\partial q_j}\right)_\beta\right)=\delta_{ij}$ and $dq_i\left(\left(\frac{\partial}{\partial p_j}\right)_\beta\right)=0$, we have
\begin{align*} \alpha_\beta(v)=\sum_{i=1}^n p_i a_i. \end{align*}
Thus $\lambda_\beta(v)=\alpha_\beta(v)$ for every $\beta\in T^*U$ and every $v\in T_\beta(T^*U)$. Therefore
\begin{align*} \lambda=\sum_{i=1}^n p_i\,dq_i. \end{align*}
[/step]
