Assume you have a function $f(x)$, which you want to approximate. As you mentioned, the first step is to write an infinite sum with unknown coefficients, which will be figured out.
$$f(x) = c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+...$$
To find $c_0$, $a$ is set to be equal to $x$.
$$f(a) = c_0+c_1(x-x)+c_2(x-x)^2+c_3(x-x)^3+...$$
Therefore, all terms including which included $(x-a)$ will now be $(x-x)$. All such terms will be ignored as they become $0$.
$$c_0 = f(a)$$
You now look at the derivative of the function $f(x)$ so we can mimic the slope. Using the Power Rule for each and every term, you reach the following.
$$f’(x) = c_1+2c_2(x-a)+3c_3(x-a)^2+...$$
To find $c_1$, $a$ is set to be equal to $x$.
$$f’(a) = c_1+2c_2(x-x)+3c_3(x-x)^2+...$$
Once again, all terms which included $(x-a)$ will be canceled.
$$f’(a) = c_1$$
You now want to look at the second derivative of the function $f(x)$ to mimic the rate at which the slope changes.
$$f’’(x) = (1\cdot 2)c_2+(2\cdot 3)c_3(x-a)+...$$
To find $c_1$, $a$ is set to be equal to $x$.
$$f’’(a) = (1\cdot 2)c_2+(2\cdot 3)c_3(x-x)+...$$
Once again, all terms which included $(x-a)$ will be canceled. This leaves you with $2c_2$.
$$f’’(a) = (1\cdot 2)c_2 \implies c_2 = \frac{f’’(a)}{1\cdot 2}$$
Now that you get the idea, you repeat the same pattern to reach the polynomial.
$$f’’’(a) = (1\cdot 2\cdot 3)c_3 \implies c_3 = \frac{f’’’(a)}{1\cdot 2\cdot 3}$$
$$f^n(a) = (n!)c_n \implies c_n = \frac{f^n(a)}{n!}$$
Therefore, you can rewrite the original polynomial with its coefficients.
$$f(x) = f(a) + f’(a)(x-a) + \frac{f’’(a)}{2!}(x-a)^2 + \frac{f’’’(a)}{3!}(x-a)^3+...$$
This general form is known as the Taylor Series. A McLaurin Series is any such series centered around $a = 0$.
Now, look at any term in the polynomial sequence.
$$\frac{f^n(0)}{n!}(x)^n$$
We can notice the following pattern which you noted. (Use the Power Rule for fractions.)
$$\frac{d}{dx}\frac{f^n(0)}{n!}(x)^n = \frac{nf^n(0)}{n!}(x)^{n-1} = \frac{f^n(0)}{(n-1)!}(x)^{n-1}$$
Compare the derivative of this $n^{th}$ term with the term before it.
$$\frac{f^{n-1}(0)}{(n-1)!}(x)^{n-1}$$
They both have the coefficient $\frac{(x)^{n-1}}{(n-1)!}$.
For integrating, the pattern is reversed. (Use the Power Rule for integrals.)
$$\int \frac{f^n(0)}{n!}(x)^n dx = \frac{f^n(0)}{(n+1)}(x)^{n+1}+C$$
Compare the integral of this $n^{th}$ term with the term after it.
$$\frac{f^{n+1}(0)}{(n+1)!}(x)^{n+1}$$
They both have the coefficient $\frac{(x)^{n+1}}{(n+1)!}$.
This answers your question about why the coefficient of the derivative of every term is the same as the coefficient of the term before it, and why the coefficient of the integral of every term is the same as the coefficient of the term after it.