In these lecture notes on Lie algebras they give a definition of the pushforward that I haven't seen before.
$$\psi_* = d\psi_x: T_xX → T_{\psi(x)}N$$
I understand $d\psi_x$, but $\psi_*$ makes no mention of which point it is attached to, so I wouldn't be able to guess its intended domain. Given a $\psi: X \to N$, its pushforward is well-defined on all the tangent spaces of $X$, I think.
We are soon asked to check that:
$$(1_X)_∗ = 1_{T_xX}$$
Again, this check only makes sense if I can compare their domains, but the object on the left doesn't reference a point. Why not? Of course, if I restrict the pushforward on the left to the space $T_xX$ I can confirm their equality, but that doesn't seem to be what this notation is suggesting.
Often, people usually just infer the domain when the point isn't explicitly mention it. But really, there are several ways of formulating the concept of push-forward. The first is point by point, and then extending to the whole tangent bundle, and the second is to define it on the entire tangent bundle, and then define it at a point via restriction.
Let $X,N$ be smooth manifolds and $\psi:X \to N$ be a smooth map.
First Approach: (Pointwise to globally)
Then, for each point $x\in X$, we can define the pushforward of $\psi$ at $x$ to be the map $\psi_{*x}: T_xX \to T_{\psi(x)}N$ defined by \begin{align} \psi_{*x} \big([\gamma]\big) := [\psi \circ \gamma] \end{align}
So, here we started by defining the pushforward at a point. We can extend this to the entire tangent bundle very easily, because given an equivalence class of curves $[\gamma] \in TX$, since the tangent spaces are all disjoint, we know that there exists a unique $x\in X$ such that $[\gamma] \in T_xX$ (in fact $x = \gamma(0)$). So, with this observation, we can define $\psi_* : TX \to TN$ as \begin{align} \psi_* \big( [\gamma]\big) &:= \psi_{*x} \big( [\gamma] \big) \quad \text{where $x$ is the unique point of $X$ such that $[\gamma] \in T_xX$} \end{align}
Second Approach: (Globally to pointwise)
(For some reason I just prefer this approach, because it seems cleaner)
Recall that the tangent bundle $TX$ is the union of all tangent spaces equipped with an appropriate differentiable structure, that is $TX := \bigcup\limits_{x \in X}T_xX$. Now, we can define the pushforward of $\psi$ to be the map $\psi_*: TX \to TN$ defined by \begin{align} \psi_*\big( [\gamma]\big) := [\psi \circ \gamma] \end{align}
Now, for each $x \in X$, denote the restriction, $\psi_* \big|T_xX$, by the more convenient symbol $\psi_{*x}$. Therefore, $\psi_{*x}$ is a map with domain $T_xX$, but the codomain is the codomain of $\psi_*$, which is $TN$. So, we get the map $\psi_{*x} : T_xX \to TN$. But now, as an easy theorem, you can prove that the image of $\psi_{*x}$ lies entirely inside the tangent space $T_{\psi(x)}N$, therefore we can without much harm, restrict the codomain as well to get a map $\psi_{*x}: T_xX \to T_{\psi(x)}N$.
Now that hopefully the notation is clear, I think their equalities are really just due to sloppiness/laziness in notation (which is pretty common in this context). These are the more proper ways of stating the equalities: $(\text{id}_X)_* = \text{id}_{TX}$ and $(\text{id}_X)_{*x} = \text{id}_{T_xX}$.
Also, in the next line the notes have a typo, and it is written in slightly lazy notation. The more pedantic way of writing it is:
The first equation is really a true equality of functions in the sense that the domains, codomains and the rules of the functions on both sides all agree. In the second equation, I was careful to specify that "for all $x \in M\dots$", and you can check for yourself, that again, the domains and codomains all match up and all the compositions are well-defined.