Taylor Remainder Theorem is
$$|R_n| = \frac{M}{n+1}|x-a|^{n+1}$$
Well, can someone explain this to me more graphically?
If $M$ is some number slightly bigger than the $(n+1)^{th}$ order of derivative of original function, well, it seem very close to finding error in Alternating Series.
$\frac{f^{n+1}(x)}{n+1}|x-a|^{n+1}$ is the $(n+1)^{th}$ term after all.
According to the integral remainder formula, if you put $h = b-a$:
$$ R_n = \frac{h^{n+1}}{n!}\int_0^1 (1-t)^n f^{(n+1)}(a+th)dt$$
Or, the Lagrange form gives you the existence of $\theta \in [0, 1]$ such that:
$$ R_n = \frac{h^{n+1}}{(n+1)!}f^{(n+1)}(a+\theta h)$$ This $\theta$ is a function of $a$ and $b$ (or $h$).
I am not sure of what your formula is supposed to mean.
You can, for instance, bound the remainder with such an M constant for all $h$ in a segment if $f \in \mathcal{C}^{n+1}$. Indeed, in such a case $f^{(n+1)}$ is continuous over any segment, thus bounded.
For a graphical explanation, you could say that the variations of a function in a neighborhood of any point cannot be much greather than its polynomial interpolation at this point, if the function is regular enough.
That means you will never see a $\mathcal{C}^{\infty}$ function going crazy like $x \mapsto x\sin(\frac{1}{x})$ around $0$:
