I'm following Ana Cannas' book, trying to prove Proposition 2.5. If you don't want to open the book, here's a sketch of the problem.
Let $X_\mu = \{ (x,\mu_x); x \in X, \, \mu_x \in T_x^*X\}$, where $X$ is a manifold. $X_\mu$ defines, thus, a submanifold in the cotangent bundle. The de Rham 1-form $\mu: X \to T^*X$ is said to depend smoothly on $x$. Define $s_\mu: X \to T^*X: x \mapsto (x,\mu_x)$ as "the 1-form $\mu$ regarded exclusively as a map", which obviously has $X_\mu$ as its image, and let $\alpha$ be the tautological 1-form on $T^*X$. Proposition 2.5 wants to prove that $s_\mu^* \alpha = \mu$.
The proof starts by noticing that one possible definition of the tautological form, $\alpha_p = (d\pi_p)^*\xi$, where $p=(x,\xi)$ and $\pi$ is the projection $\pi:T^*X \to X: (x,\xi) \mapsto x$, says that if $p=(x,\mu_x)$, then $\alpha_p = (d\pi_p)^*\mu_x$ (pretty obvious). She then writes that $(s_\mu^* \alpha)_x = (ds_\mu)_x^* \alpha_p = \dots$. I can't understand how $s_\mu^*$ became $(ds_\mu)^*$.
If $s_\mu: X \to T^*X:x \mapsto (x,\mu_x)$, then $s_\mu^*: T^* X \to T^*(T^*X): (x,\gamma) \mapsto (?,?,?)$, and I also can't get what $(ds_\mu)^*$ is/does. I think I'm having problems visualising the actions of all those mappings.
The definition of $(s_{\mu}^*\alpha)_x$ is $(ds_{\mu})^*_x{\alpha_p}$. I think the notation may be a little overwhelming. Note that this "$d$" is not an exterior derivative, it is just the derivative of $s_{\mu}$ as a map. The first star is the pull-back of the $1$-form $\alpha$, the second star is the (linear-level) adjoint of the derivative (which is morally the same as the pull-back, but is happening only on the linear level).
To be quite explicit, let's purge the notation abuse: let $f: M \to N$ be a smooth function and $\alpha$ be a $1$-form on $N$. Let $(f_*)_p: T_pM \to T_{f(p)}N$ denote the derivative at $p$, and let $L^T$ denote the adjoint of a linear map $L:V \to W$ (i.e., the map $L^T:W^* \to V^*$ given by $L^T(\tau)=\tau \circ L$). Then, by definition: $$ (f^*\alpha)_p:=\big((f_*)_p \big)^T(\alpha_{f(p)}).$$ Now you can translate the above equality back to your needs.