We need to find all functions $f: [0, \infty) \to [0, \infty)$ that satisfy the following conditions:
1. $f(2) = 0$.
2. $f(x) \neq 0$ for $0 \leq x < 2$.
3. $f(xf(y))f(y) = f(x+y)$ for all $x, y \geq 0$.

Let's begin by analyzing these conditions:

1. **Condition $f(2) = 0$:** According to this condition, $f(x) = 0$ when $x \geq 2$.

2. **Functional Equation $f(xf(y))f(y) = f(x+y)$:** For $x, y \geq 0$.

To explore this, assume $x < 2$ and $y = 2$. Substituting into the functional equation, we have:
$$f(xf(2))f(2) = f(x + 2).$$
Since $f(2) = 0$, this simplifies to:
$$0 = f(x + 2).$$
Therefore, $f(x) = 0$ for all $x \geq 2$.

3. **Explore values in the domain $0 \leq x < 2$:**

According to the problem, on $0 \leq x < 2$, $f(x) \neq 0$. Let's assume:
$$f(x) = \frac{2}{2-x}.$$
We will verify if this function satisfies the given functional equation. Calculate:
$$f(xf(y)) = f\left(x \cdot \frac{2}{2-y}\right) = f\left(\frac{2x}{2-y}\right).$$
This becomes clear when $x < 2$ and $y < 2$. Substitute back into the functional equation:
$$f\left(\frac{2x}{2-y}\right) \cdot \frac{2}{2-y} = f(x+y).$$
If $f(x) = \frac{2}{2-x}$, then:
$$f\left(\frac{2x}{2-y}\right) = \frac{2}{2 - \frac{2x}{2-y}} = \frac{2(2-y)}{2(2-y) - 2x} = \frac{2(2-y)}{4 - 2y - 2x}.$$
Simplifying, we have:
$$= \frac{2(2-y)}{2(2-x-y)} = \frac{2}{2-x-y},$$
confirming that the left-hand side becomes:
$$\frac{2}{2-x-y} \cdot \frac{2}{2-y} = \frac{2}{2-(x+y)} = f(x+y).$$
Hence, this function satisfies the functional equation for $0 \leq x, y < 2$.

Thus for $f: [0, \infty) \to [0, \infty)$ defined as:
$$f(x) = \begin{cases} \frac{2}{2-x}, & 0 \leq x < 2, \\ 0, & x \geq 2, \end{cases}$$
this indeed satisfies all the required conditions.

Therefore, the function is:
$$\boxed{ f(x) = \begin{cases} \frac{2}{2-x}, & 0 \leq x < 2, \\ 0, & x \geq 2. \end{cases} }$$

This concludes our construction of such function $f$.