What is the average height of a binary tree?
There is no reason to believe that both definitions describe the same measure. You can write $\operatorname{avh}_1$ recursively, too:

$\qquad \displaystyle \operatorname{avh}_1(N(l,r)) = \frac{\operatorname{lv}(l)(\operatorname{avh_1}(l) + 1) + \operatorname{lv}(r)(\operatorname{avh_1}(r) + 1)}{\operatorname{lv}(l) + \operatorname{lv}(r)}$

with $\operatorname{avh}_1(l) = 0$ for leaves $l$. If you don't believe that this is the same, unfold the definition of $\operatorname{avh}_1$ on the right hand side, or perform an induction proof.

Now we see that $\operatorname{avh}_1$ works quite differently from $\operatorname{avh}_2$. While $\operatorname{avh}_2$ weighs the recursive heights of a nodes children equally (adding and dividing by two), $\operatorname{avh}_1$ weighs them according to the number of leaves they contain. So they are the same (modulo the anchor) for leaf-balanced trees, that is balanced in the sense that sibling trees have equally many leaves. If you simplify the recursive form of $\operatorname{avh}_1$ with $\operatorname{lv}(l) = \operatorname{lv}(r)$ this is immediately apparent. On unbalanced trees, however, they are different.

Your calculations are indeed correct (given your definition); note that the example tree is not leaf-balanced.