Here you can do a separation of the cases because there are very few. Here we go:
$X = 0; Y = 0 \Rightarrow |X - Y| = 0;(X+Y) = 0;(X+Y)|X-Y| = 0$
$X = 1; Y = 0 \Rightarrow |X - Y| = 1;(X+Y) = 1;(X+Y)|X-Y| = 1$
$X = 0; Y = 1 \Rightarrow |X - Y| = 1;(X+Y) = 1;(X+Y)|X-Y| = 1$
$X = 1; Y = 1 \Rightarrow |X - Y| = 0;(X+Y) = 2;(X+Y)|X-Y| = 0$
By giving a $1/4$ weight to each of these cases, you should find that the expected value of $E\left((X+Y)|X-Y|\right)$ is $1/2$ and not $0$. But the covariance is
$$E\left((X+Y)|X-Y|\right) - E(X+Y)E(|X-Y|) = 1/2 - 1 \cdot 1/2 = 0.$$
Short/Technical Answer:
It's untrue that $cov(y,z) = 0$ whenever $z$ does not directly affect $y$. The key is the word "directly". We hope to find an instrument such that "z influences x which influences y". This would mean that $cov(y,z|x) = 0$ (that is, if we observe the value of $x$, knowing $z$ tells us nothing incremental about $y$), but it does not mean that $cov(y,z) = 0$ unconditional on $x$.
Long/Intuitive Answer:
The distinction you need to make here is the difference between an associative (or correlative) effect and a causal one.
Let's first look at an example (linked at bottom). Suppose you want to know how the size of a political protest ($x$) causally affects local political outcomes ($y$).
However, if you simply look at the association between $x$ and $y$, you may note a number of potential issues. For example, the size of the protest ($x$) could itself be correlated with other factors like the importance of the issue or the likelihood of the protest causing change. These factors are also correlated with $y$. Hence, they confound the relationship between $x$ and $y$.
So how can an instrument variable help us? Consider the instrument of how good the weather was on the day of the protest ($z$). The intuition is that weather ($z$) could directly impact crowd size ($x$); however weather on the day of a protest ($z$) should not directly influence a local political outcome ($y$). The only way you would expect $z$ and $y$ to be related is through $x$.
Example Source: https://dash.harvard.edu/bitstream/handle/1/13457753/TeaParty_Protests.pdf
Best Answer
The standard formula does work, just needs a bit of manipulation
$$Cov(y,z) = E(yz) - E(y)E(z) = E(yz\mid z=1)P(z=1) - E(y)p$$
$$[E(y\mid z=1) - E(y)]p = \Big[E(y\mid z=1) - \big[E(y|z=1)p + E(y\mid z=0)(1-p)\big]\Big]p$$
$$=\Big[ E(y \mid z=1)(1-p) - E(y \mid z=0)(1-p) \Big]p$$
$$=\Big[ E(y \mid z=1) - E(y \mid z=0) \Big]p(1-p)$$