### Solution:

#### Part (1) Proof:

Given that $\overrightarrow{m}=({a+c,a})$ and $\overrightarrow{n}=({a-c,b})$ are perpendicular, we have:
$$\overrightarrow{m} \cdot \overrightarrow{n} = (a+c)(a-c) + ab = a^2 - c^2 + ab = 0$$
From the cosine rule, we know that:
$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$
Substituting $a^2 - c^2$ from the dot product equation, we get:
$$2ab\cos C - b^2 + ab = 0$$
Rearranging, we find:
$$2a\cos C - b + a = 0$$
Using the Law of Sines, $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$, we can rewrite the equation as:
$$2\sin A\cos C - \sin B + \sin A = 0$$
Since $\sin B = \sin (A+C)$, we have:
$$2\sin A\cos C - \sin (A+C) + \sin A = 0$$
Simplifying using trigonometric identities, we get:
$$\sin (A-C) = \sin (-A)$$
Given $A$ and $C$ are interior angles of the triangle, we conclude:
$$A - C = -A \implies C = 2A$$
Thus, we have proven that $C = 2A$.

#### Part (2) Finding the Range:

Using the Law of Sines and the fact that $C = 2A$, we express $\frac{b}{a} + \left(\frac{2a}{c}\right)^2$ as:
$$\frac{\sin B}{\sin A} + \frac{4\sin^2 A}{\sin^2 C} = \frac{\sin (180^\circ - 3A)}{\sin A} + \frac{4\sin^2 A}{\sin^2 2A}$$
Simplifying using trigonometric identities, we get:
$$\frac{\sin 3A}{\sin A} + \frac{4\sin^2 A}{4\sin^2 A \cos^2 A} = \cos 2A + 2\cos^2 A + \frac{1}{\cos^2 A}$$
Further simplification gives:
$$4\cos^2 A + \frac{1}{\cos^2 A} - 1$$
Given $A \in \left(\frac{\pi}{6}, \frac{\pi}{4}\right)$, we find $\cos^2 A \in \left(\frac{1}{2}, \frac{3}{4}\right)$. Letting $t = \cos^2 A$, the expression becomes:
$$4t + \frac{1}{t} - 1, \quad t \in \left(\frac{1}{2}, \frac{3}{4}\right)$$
Analyzing the function $y = 4t + \frac{1}{t} - 1$ over the interval $\left(\frac{1}{2}, \frac{3}{4}\right)$, we find it is monotonically increasing. Therefore, the range of values for $\frac{b}{a} + \left(\frac{2a}{c}\right)^2$ is:
$$\boxed{(3, \frac{10}{3})}$$