Why are polynomials tangential to the $x$ axis at real double roots?

First a comment

Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.

Regarding your question

So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get

$$p^\prime(x) = 2(x-a)q(x) + (x-a)^2 q^\prime(x).$$

Therefore you have $p(a)=p^\prime(a)=0$. Proving that the graph of the function $x \ \mapsto p(x)$ is tangent to the $x$-axis at $x=a$.

You can prove this easily using calculus, but I'll give an even more elementary argument.

The graph of $f(x)=(x-a)^{2}$ is an upward facing parabola. In particular, it is the graph of $f(x)=x^{2}$ shifted to the right by $a$ units (if a is positive).

Note that, being a square, $(x-a)^{2}$ is always greater than or equal to $0$. And it is only equal to $0$ at $x=a$.

Combining the above two observations, the graph of $f(x)$ is tangent to the $x$-axis at $x=a$.