A non-path-dependent American derivative pays an amount $g(s)$ depending only on the current stock price $s$ whenever it is exercised.
The value at a current time is the maximum of the value of immediate exercise and the value of delayed exercise; the latter is defined recursively as before.
So if you have a choice between \$5 and a derivative valued at \$4, you should always choose the \$5, because in theory you could use \$4 to buy the derivative immediately and put the other \$1 in the bank. That is, we are assuming trading that we do is instantaneous whereas stock movements and interest accrual are not. If the stock could rise sharply while we are choosing the \$5, the logic would be different.

A path-dependent American derivative is really no different: we again price the derivative at any time as the maximum of the value of instantaneous exercise and the value of delayed exercise.

Note that the notion of optimal exercise of an option is straightforward, to determine it is merely a matter of choosing the largest of two numbers.

A stopping time in the $N$-period binomial model is a random variable taking integer values from $0$ to $N$ or $\infty$, and whose value $\tau(\omega)$ only depends on $\omega_1\cdots\omega_{\tau(\omega)}$. In other words, any choice of the remaining bits of $\omega$ gives the same value to $\tau$.

The notion of a stopping time can be used to express the price of an American derivative without talking about maxima of two numbers and recursion.
Namely, the price $V_n$ is the maximum over all stopping times (i.e., over all option exercise algorithms) taking values $\ge n$, of
$$
\tilde{\mathbb E}\left( 1_{\tau\le N} G_\tau/(1+r)^{\tau-n}\right)
$$
Here $G_n$ is the value of the derivative if exercised after $n$ steps.

In other words, price the derivative by the discounted risk-neutral expected payoff according to the exercise algorithm that maximizes it.

To justify the price formula, note that (1) once $\tau$ is fixed the option is equivalent to a European one, if we put payoff in the money market after exercise, and (2) if we assume that prices for options exist at every place in the tree then the optimal $\tau$ strategy is just to choose optimally (to exercise or not) at each stage.

We can note that a call option tends to gain value as time goes by, since we get to buy for a smaller and smaller amount in discounted terms. A put option tends to lose value, because being allowed to sell for 100 dollars is not so desirable in the future when 100 dollars is a smaller amount in discounted terms. Therefore a call option should be exercised at time $N$ (and so American call = European call, in effect). The textbook proves this by going via convex functions (a call option has a convex payoff function, and Jensen’s inequality for convex functions is used) but we can consider the call option case directly: $g(s)=(s-K)^+$. We want to show that
$$
\max_{\tau\in\mathcal S_n}\tilde{\mathbb E}\left( 1_{\tau\le N} g(S_\tau)/(1+r)^{\tau-n}\right) = \tilde{\mathbb E}\left( S_N^+ / (1+r)^N \right)
$$

Let us assume $S_n>K$ and compare exercise at time $n$ with waiting one more step.
Case 1: $uS_n$ and $dS_n$ are both $>K$. Then
$$
(1+r)^{-1} \tilde{\mathbb E}(S_{n+1}-K)^+ = (1+r)^{-1} ((1+r)S_n – K) = S_n – \frac{K}{1+r} \ge S_n – K.
$$
Case 2: Otherwise, then $dS_n<K$. We may assume $S_n> K$. At the next step we $\tilde{\mathbb E}$-expect the stock price to be at $(1+r)S_n$ and so
it is enough to check that
$$
(1+r)^{-1}\tilde{\mathbb E}(S_n-K)^+ = (1+r)^{-1} (uS_n – K) \tilde{\mathbb P}(H) = (1+r)^{-1}(uS_n-K)\frac{1+r-d}{u-d} \ge^{!} S_n – K
$$
We proceed as follows:
$$
S_n-K \le \frac{uS_n-K}{1+r}\cdot\frac{1+r-d}{u-d}
$$
$$
(1+r)(u-d)(S_n-K) \le (uS_n-K)(1+r-d)
$$
$$
(1+r)(u-d)(S_n-K) \le (uS_n-K)(1+r-d)
$$
$$
[(u-d-1)r+(u-1)]K \ge (u-1-r)dS_n
$$
For this it suffices, since $dS_n\le K$, that
$$
(u-d-1)r+(u-1) \ge u-1-r
$$
This is true since $u\ge d$.