Is covariance transitive?

If given:

$$cov(x, y) = 0, cov(x, z) = 0$$

then can we conclude that:

$$cov(y, z) = 0$$

There is a kind of stupid counter-example. If $$y = z$$ then your conditions degenerate to one condition, and are asking if $$\text{cov}(x,y)=0 \implies \text{cov}(y,y)=0$$, but this certainly need not be true!
No. If $$X, Y$$ are independent and $$Y, Z$$ are independent, then $$X, Z$$ does not have to be independent. For example, if $$X=0$$ with probability $$1$$, then $$X, Y$$ are independent and $$Y, Z$$ are independent, but $$Y$$ and $$Z$$ are arbitrary, and can depend on each other.