A fixed point of a function $f$ is a number $t$ where $f(t)=t$. Geometrically, this means that the graph of $f$ crosses the line $y=x$ at the point $x=t$. Similarly, the graphs of $f\circ f$ and $f\circ f\circ f$ etc. also across the line $y=x$ when $x=t$.
How do we conclude linear functions at most can have one fixed point (with the exception of $f(x)=x$) while quadratics have at most two?
I have plotted an example in the case $f(x)=x^2-2$:
If $f(x) = ax + b$ is linear function, then equation $f(x) = x$ can be rearranged to $$ ax + b = x \iff (a - 1)x = -b.$$ If $a\neq 1$, then we have unique solution $x = \frac b{1-a}$, while when $a = 1$ we either have infinitely many solutions when $b = 0$ (exception mentioned in the comments) to equation $0x = 0$ or no solutions when $b \neq 0$ to equation $0x = -b$.
When $f(x)$ is polynomial of degree $n \geq 2$, then equation $f(x) = a_n x^n + a_{n-1}x^{n-1} + \ldots + a_1 x + a_0 = x$ where $a_n \neq 0$ can be rearranged to $$ a_n x^n + a_{n-1}x^{n-1} + \ldots + a_2 x^2 + (a_1-1) x + a_0 = 0, $$ so fixed points of $f(x)$ are precisely the roots of polynomial (of degree $n$) $g(x) = f(x) - x$. According to fundamental theorem of algebra, such polynomial can have at most $n$ real roots, so $f$ has at most $n$ fixed points.
The case $n = 1$ was special, because degree of $f(x) - x$ could be different than degree of $f(x)$.