Write $f(x)=f_1(x)+f_2(x)$ where $f_1(x)=(|x|+|x-1|)\sin x$ and $f_2(x)=[x\sin x]$.

Analysing $g(x)=x\sin x$ on $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$: $g$ is even, $g(0)=0$, $g(\pm \pi /2)=\pi /2\approx 1.57$, and $g^{\prime }(x)=\sin x+x\cos x$ is non-negative on $[0,\pi /2)$, making $g$ increasing there. So $g$ takes the value $1$ exactly once in $(0,\pi /2)$, say at $x_1$, and by symmetry also at $-x_1$. The range of $g$ on the interval is $[0,\pi /2)$.

Behaviour of $f_2=[g(x)]$: $[g(x)]$ jumps at points where $g$ crosses an integer. The only integer values $g$ attains in $[0,\pi /2)$ are $0$ (at $x=0$) and $1$ (at $\pm x_1$). At $x=0$, $g$ is locally $0$, so no jump occurs there. At $x=\pm x_1$, $g$ crosses $1$, producing a jump discontinuity. So $f_2$ is discontinuous (and hence non-differentiable) at exactly $\{x_1,-x_1\}$.

Behaviour of $f_1$: $f_1$ is continuous everywhere on the interval. Its only non-differentiable points are where $|x|$ or $|x-1|$ kinks, i.e. $x=0$ and $x=1$. The function $\sin x$ vanishes at $x=0$, which smooths the kink there; so $f_1$ is differentiable at $0$ but not at $x=1$.

Note $x_1\ne 1$ since $g(1)=\sin 1\approx 0.84\ne 1$.

Counting discontinuities ($\alpha$): Only $f_2$ contributes, and $f_1$ is continuous everywhere, so the discontinuities of $f$ are exactly $\{x_1,-x_1\}$. Hence $\alpha =2$.

Counting non-differentiable points ($\beta$): At $x_1$ and $-x_1$, $f_2$ is non-differentiable while $f_1$ is differentiable, so $f$ is non-differentiable. At $x=1$, $f_1$ is non-differentiable while $f_2$ is differentiable (locally constant). At $x=0$, both are differentiable. So $\beta =3$.

$\therefore \alpha +\beta =2+3=5$.