Is there a bound for Lipschitz constant in terms of second differences? It is easy to show that if $f\colon[0,1]\to\mathbb R$ and $|f|\leq A$ and $|f''|\leq B$ then~$|f'|\leq 4A+B$. Indeed, by Taylor formula with remainder $f(x)=f(c)+(x-c)f'(c)+\frac12(x-c)^2f''(d)$ where $d$ is between $x$ and $c$. Therefore $f'(c)=\frac{f(x)-f(c)-\frac12(x-c)^2f''(d)}{x-c}$. Now let $c$ be a maximum of $|f'|$. We can assume without loss of generality that $c\geq\frac12$ (otherwise apply the same argument to the function $f(\frac12-x)$ to get the same bound). Then  \begin{aligned} |f'(c)|&\leq2\left|f(0)-f(c)-\frac12(0-c)^2f''(d)\right| \\&\leq2|f(0)|+2|f(c)|+c^2 |f''(d)| \\&\leq4A+B \end{aligned}  as required.  In particular, $f$ is $(4A+B)$-Lipschitz.
The question is, can one obtain similar bounds for the Lipschitz constant of $f$ if one does not require differentiability but only assumes a bound on second differences: $|f(x)-2f(x+h)+f(x+2h)|\leq Bh^2$ ? 
(asked unsuccessfully at MSE.)
 A: Smoothifying by convolution as Pietro Majer suggests is pretty ok, but if you prefer more direct argument, you may use a standard
Lemma. If a bounded function $f$: $[0,1]\to \mathbb{R}$ satisfies $f(\frac{x+y}2)\leqslant \frac{f(x)+f(y)}2$, then $f$ is convex. 
Proof. At first, we prove that $f$ is continuous on $(0,1)$. If not, there exists a $c\in (0,1/2)$ and a sequence of points $x_n\in [c,1-c]$ and $\delta_n\rightarrow 0$ such that $f(x_n+\delta_n)-f(x_n)\geqslant c$. We get $f(x_n+k\delta_n)\geqslant f(x_n)+kc$ whenever $0\leqslant x_n+k\delta_n\leqslant 1$, this contradicts to the assumption that $f$ is bounded when $\delta_n$ tends to 0. The rest is easy: we get $f(\alpha x+(1-\alpha)y)\leqslant \alpha f(x)+(1-\alpha)f(y)$ for $x\ne y\in [0,1]$ and $\alpha\in \{1/2^n,2/2^n,\dots,(2^n-1)/2^n\}$ by induction in $n$. For arbitrary $\alpha\in (0,1)$ approximate it by such numbers and use continuity (at a point $\alpha x+(1-\alpha)y\in (0,1)$).
Thus your condition implies that the function $g(x)=f(x)-Bx^2/2$ is concave and $h(x)=f(x)+Bx^2/2$ is convex. Fix $0<a<b<1$, we want to estimate $|\frac{f(b)-f(a)}{b-a}|$. By replacing $f(x)$ to $f(1-x)$, we may assume that $b\geqslant 1/2$. Next, by replacing $f$ to $-f$, we may estimate $\frac{f(b)-f(a)}{b-a}$ from below. Ok, we have $$\frac{f(b)-f(a)}{b-a}+B\frac{a+b}2=\frac{h(b)-h(a)}{b-a}\geqslant \frac{h(b)-h(0)}{b}\geqslant -2\frac Ab+\frac {Bb}{2},$$
$$\frac{f(b)-f(a)}{b-a}\geqslant-2\frac{A}b-B\frac{a}2\geqslant -2\frac{A}b-B\frac{b}2\geqslant -\max(4A+\frac{B}4,2A+\frac{B}2).$$
