It is common practice to write:
$$f(x+h) = f(x) + f'(x)h + O(h^2)$$
We know that the remainder has the form $ \frac{1}{2}f''(\xi)h^2 $ with $\xi \in (x, x+h)$. That is, $\xi$ is dependent of $h$.
What is the justification to use the big-$O$ notation, that implies there exists $C > 0 $, independent of $h$ so that
$$f(x+h) \leq f(x) + f'(x)h + Ch^2$$
On
Recall that this writing is valid on the assumption that $f$ is twice continuously differentiable. The correspondence $g: h\mapsto f''(\xi_h)$ defines a function. Now:
$$| f(x+h) -f(x) - f'(x)h| = \left| \frac{g(h)}{2}\right|h^2$$
By continuity of $f''$, $g(h) \to f''(x)$ as $h\to 0$, so there exists $\delta > 0$ such that for $|h| < \delta$, $|g(h) - f''(x)| < 1$, thus:
$$\forall |h| < \delta, | f(x+h) -f(x) - f'(x)h| < \left( \frac12 (1 + |f''(x)|) \right) h^2$$
Yes it is. The justification is the remainder estimate from Taylor's theorem, namely that if you have an $n+1$ times differentiable function, then
$$f(x) = T_n(x) +R_n(x)$$
where $T_n(x)$ is the $n^{th}$ order Taylor polynomial and the remainder has the bound
for some $\xi$ on the interval of consideration. Expand the $(x-\xi)^{n+1}$ using the binomial theorem to get your desired bound.