Smoothifying by convolution as Pietro Majer suggests is pretty ok, but if you prefer more direct argument, you may use a standar

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 $f(x_n+\delta_n)-f(x_n)\geqslant c$. Replacing if necessary $f(x)$ to $f(1-x)$ we may suppose that $\delta_n>0$ for infinitely many $n$. We get $f(x_n+k\delta_n)\geqslant f(x_n)+kc$ whenever $x_n+k\delta_n<1$, 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 on $n$. 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 replace 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 $$\frac{f(b)-f(a)}{b-a}+B\frac{a+b}2=\frac{h(b)-h(a)}{b-a}\geqslant \frac{h(b)-h(0)}{b}\geqslant -2\frac Ab+\frac {Bb}{2},$$ $$\frac{f(b)-f(a)}{b-a}\geqslant-2\frac{A}b-B\frac{a}2\geqslant -2\frac{A}b-B\frac{b}2\geqslant -\max(4A+\frac{B}4,2A+\frac{B}2).$$

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 $$\frac{f(b)-f(a)}{b-a}+B\frac{a+b}2=\frac{h(b)-h(a)}{b-a}\geqslant \frac{h(b)-h(0)}{b}\geqslant -2\frac Ab+\frac {Bb}{2},$$ $$\frac{f(b)-f(a)}{b-a}\geqslant-2\frac{A}b-B\frac{a}2\geqslant -2\frac{A}b-B\frac{b}2\geqslant -\max(4A+\frac{B}4,2A+\frac{B}2).$$

Smoothifying by convolution as Pietro Majer suggests is pretty ok, but if you prefer more direct argument,

note that 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. Therefore for $0<a<b<1$ we have $$\frac{f(b)-f(a)}{b-a}+B\frac{a+b}2=\frac{h(b)-h(a)}{b-a}\geqslant \frac{h(b)-h(0)}{b}\geqslant -2\frac Ab+\frac {Bb}{2},$$ $$\frac{f(b)-f(a)}{b-a}\geqslant-2\frac{A}b-B\frac{a}2\geqslant -2\frac{A}b-B\frac{b}2\geqslant -\max(4A+\frac{B}4,2A+\frac{B}2)$$ provided that $b\in [1/2,1)$, which still may be supposed.

Smoothifying by convolution as Pietro Majer suggests is pretty ok, but if you prefer more direct argument, you may use a standar

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 $f(x_n+\delta_n)-f(x_n)\geqslant c$. Replacing if necessary $f(x)$ to $f(1-x)$ we may suppose that $\delta_n>0$ for infinitely many $n$. We get $f(x_n+k\delta_n)\geqslant f(x_n)+kc$ whenever $x_n+k\delta_n<1$, 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 on $n$. 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 replace 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 $$\frac{f(b)-f(a)}{b-a}+B\frac{a+b}2=\frac{h(b)-h(a)}{b-a}\geqslant \frac{h(b)-h(0)}{b}\geqslant -2\frac Ab+\frac {Bb}{2},$$ $$\frac{f(b)-f(a)}{b-a}\geqslant-2\frac{A}b-B\frac{a}2\geqslant -2\frac{A}b-B\frac{b}2\geqslant -\max(4A+\frac{B}4,2A+\frac{B}2).$$