The basic intuition is that in a neighbourhood of $0$, $f$ is positive. This leads to $f'$ being increasing (as $f'' > f$), and thus $f$ is increasing. If you picked the "leftmost" point where $f$ is 0, then in that interval, you need $f'$ to become $0$ at some point.
To formalize this, an elementary proof:
Since $f(0) = 0$, and $f'(0) \gt 0$, there is a $\delta \gt 0$ such $f(x) \gt 0$ for all $x \in (0, \delta)$.
This we can see by using the $\epsilon-\delta$ definition of derivative and choosing $\epsilon = \frac{f'(0)}{2}$.
Now assume there is some point $y \gt 0$ where $f(y) \le 0$. This implies there is some point $y' \gt 0$ such that $f(y') = 0$
Now let $S$ be the set defined by $S = \{ y: y \ge \delta, f(y) = 0\}$.
Since S is bounded below, $c = \inf S$ (greatest lower bound) is well defined. By continuity of $f$ we have that $f(c) = 0$ (we can pick a sequence $c_n \to c$ and $c_n \in S$). Note that $c \ge \delta \gt 0$.
Now for any $0 \lt x \lt c$, we have that $f(x) \gt 0$. This is because, for $x \lt c$, we cannot have $f(x) = 0$ (as $c = \inf S$), and if $f(d) \lt 0$ for some $d$, then by continuity, there is a point $e \lt d \lt c$ such that $f(e) = 0$.
Thus for $x \in (0,c)$, we have $f(x) \gt 0$ and $f(0) = f(c) = 0$.
In this interval $f''(x) \ge f(x) \gt 0$. Thus $f'(x)$ is increasing, and since $f'(0) \gt 0$, we have that $f'(x) \gt 0$ for all $x \in (0,c)$.
But since $f(0) = f(c)$, we must have that $f'(\eta) = 0$ for some $\eta \in (0,c)$, by Rolle's theorem.
A contradiction.
Thus there is no $y \gt 0$ for which $f(y) \le 0$.
The inequality at which you arrived implies that $|f'(x) - f'(y)| \leq M$ for all $0 < x,y < 1$. If $f'$ is unbounded on $]0,1[$, then for every $B > 0$ there are some $0 < x,y < 1$ such that $|f'(x) - f'(y)| > B$; hence $f'$ is bounded on $]0,1[$ from above by $M$. Then mean-value theorem gives that for every pair of $0 < x,y < 1$ there is some $x < \xi < y$ such that $|f(x)-f(y)| \leq |f'(\xi)||x-y| \leq M|x-y|$. If $\varepsilon > 0$, the number $\delta := \varepsilon/M$ is such that $|x-y| < \delta$ implies $M|x-y| < \varepsilon$; hence $f$ is uniformly continuous on $]0,1[$.
Best Answer
Given that $f''\geq0$, $f$ is convex. This implies that $f$ lies weakly above its tangent line at any point. Formally, for any $x_0\in\mathbb R$, $$f(x)\geq f(x_0)+f'(x_0)(x-x_0).$$ Now, if $f'(x_0)>0$, then the right-hand side goes to infinity as $x\to\infty$, which contradicts $f$ being bounded. Similarly, if $f'(x_0)<0$, then the right-hand side diverges to $\infty$ as $x\to-\infty$, which implies that $\lim_{x\to-\infty}f(x)=\infty$, another contradiction. It follows that $f'(x_0)=0$. Since this is true for any $x_0\in\mathbb R$, $f'=0$, so that $f$ is constant.