The function $f(x) = |[x]| + \sqrt{x-[x]}$ is composed of two parts : the greatest integer function $[x]$, and the fractional part function $x-[x]$.

1. The greatest integer function $[x]$ is discontinuous at every integer, since it jumps from one integer to the next without taking any values in between. The absolute value does not affect where this function is continuous or discontinuous. So within the interval $(-2,1)$, $[x]$ is discontinuous at $-2, -1, 0$ and $1$. However, because the interval is open $(-2,1)$, the endpoints $-2$ and $1$ are not included.

2. The function $\sqrt{x-[x]}$ is the square root of the fractional part of $x$. The fractional part function $x-[x]$ is continuous everywhere, as it always takes a value between $0$ and $1$ (inclusive of $0$, exclusive of $1$). However, the square root function $\sqrt{x}$ is only defined for $x \geq 0$, and so $\sqrt{x-[x]}$ is discontinuous wherever $x-[x] < 0$. This happens exactly at the points where $x$ is a negative integer, as the fractional part of a negative integer is $1$ (considering that the "fractional part" is defined as the part of the number to the right of the decimal point, which for negative numbers works a bit differently). Within the interval $(-2,1)$, this is the case for $-2$ and $-1$. However, since the interval is open $(-2,1)$, the endpoint $-2$ is not included.

Now, let's analyze the discontinuities within the given interval $(-2,1)$:

At $x=-1$: $f(-1^{+})=1+0=1$ and $f(-1^{-})=2+1=3$. Since these two values are different, $f(x)$ is discontinuous at $x=-1$.

At $x=0$: $f(0^{+})=0+0=0$ and $f(0^{-})=1+1=2$. Again, these two values are different, so $f(x)$ is discontinuous at $x=0$.

At $x=1$: $f(1^{+})=1+0=1$ and $f(1^{-})=0+1=1$. These two values are the same, so $f(x)$ is continuous at $x=1$. However, this point is not within the open interval $(-2,1)$.

So within the interval $(-2,1)$, the function $f(x) = |[x]| + \sqrt{x-[x]}$ is discontinuous at the points $-1$ and $0$ (2 points in total).