To find the minimum value of $\frac{a+b}{ab}$ given $a > 0$, $b > 0$, and $a+2b=2$, we proceed as follows:

1. First, we express $\frac{a+b}{ab}$ in terms of the inverses of $a$ and $b$:

$$\frac{a+b}{ab} = \frac{1}{a} + \frac{1}{b}$$

2. Recognizing that $a+2b=2$, we can multiply the sum of inverses by a clever form of $1$ (which is $\frac{a+2b}{2}$) to facilitate the use of the AM-GM inequality:

$$\frac{1}{a} + \frac{1}{b} = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right)(a+2b)$$

3. Simplify the expression by distributing and rearranging terms:

$$= \frac{1}{2}\left(3 + \frac{2b}{a} + \frac{a}{b}\right)$$

4. Apply the AM-GM inequality, $\sqrt{xy} \le \frac{x+y}{2}$, to the terms $\frac{2b}{a}$ and $\frac{a}{b}$:

$$\frac{1}{2}\left(3 + \frac{2b}{a} + \frac{a}{b}\right) \ge \frac{1}{2}\left(3 + 2\sqrt{\frac{2b}{a} \cdot \frac{a}{b}}\right)$$

5. Simplify the expression under the square root and the inequality:

$$= \frac{1}{2}\left(3 + 2\sqrt{2}\right)$$

6. The equality holds (which means we have found the minimum) if and only if $\frac{2b}{a} = \frac{a}{b}$, or equivalently $a^2 = 2b^2$. Solving this together with $a+2b=2$ gives $b = 2 - \sqrt{2}$ and $a = 2\sqrt{2} - 2$.

Therefore, the minimum value of $\frac{a+b}{ab}$, given the conditions, is $\boxed{\frac{1}{2}(3+2\sqrt{2})}$.