Assume $f(x)$ is twice differentiable and $|f''(x)|\leq M$ for all $x\in R$.
Fix $a \in R$. Use the Mean Value Theorem to show that for any $x \in R, x\neq a$, there is $c \in (x, a)$, such that $$ f(x)-L_a(x)=(f'(c)-f'(a))(x-a) $$ and $$ |f(x)-L_a(x)|\leq M(x-a)^2. $$
Here $L_a$ is the linearization of $f$ at $a.$
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$\begingroup$One simply uses the function $g(x) = f(x) - L_a(f, x)$. We know $g$ is differentiable if $f$ is differentiable. So then by the mean value theorem, there is a $c$ so that $$ \frac{g(x) - g(a)}{x - a} = g'(c) = f'(c) - f'(a),$$ which rearranges into your desired equality.
For the second part, note that $f'(c) - f'(a) = f''(\xi)(c - a)$ for some $\xi \in (a, c)$, again by the mean value theorem. As $f'' \leq M$, we then know that $f'(c) - f'(a) \leq M(c-a)$.
$\endgroup$ $\begingroup$Hint: Define $g(x) = f(x) -L_a(x).$ Apply the MVT to $g(x) - g(a).$
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