Attempt towards a counter-example:
Consider $f:[0,1]\to \mathbb{R}$, $f(x)=-(x-\frac{1}{2})^4$.
$f$ takes its largest value at $x=0.5 \in [0,1]^o$, i.e. the interior of $[0,1]$.
$f''(x)=-12(x-\frac{1}{2})^2$, and $|f''(x)|\leq 3$, $\forall x \in [0,1]$.
Now, $|f''(0)|=|f''(1)|=3$. Now, $|f''(0)|+|f''(1)|=6\nleq3$
Does this example work?
Kindly VERIFY
Your counterexample is correct. The statement is (trivially) correct if $a \ge 2$ and wrong if $a < 2$. As a counterexample one can choose any twice-differentiable function $f: [0, a] \to \Bbb R$ which has a maximum in the interior of the interval and where $f''$ attains its maximum both at $x=0$ and $x=a$. So another choice would be $f(x) = -(x-a/2)^2$.
On the other hand, one can prove that $|f'(0)|+|f’(a)|\leq am$, using the fact that $f'(c) = 0$ at the point $c$ where $f$ attains its maximum.