Your equation $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}$ is actually an equation relating to vector addition, $\vec v_{\rm ac} =\vec v_{\rm ab} + \vec v_{\rm bc}$
In one dimension this equation could be written as $v_{\rm ac}\,\hat x =v_{\rm ab} \,\hat x + v_{\rm bc}\,\hat x$ where $v_{\rm ac}, \,v_{\rm ab}$ and $v_{\rm bc}$ are components of the velocity in the $\hat x$ direction.
You can think of the $\hat x$ direction as the direction that you have chosen to be positive.
Going back to your equation $v_{\rm ac} =v_{\rm ab} + v_{\rm bc}$ and noting that these are components in a positive direction you have chosen it is not difficult to assign signs.
Suppose that you chose West as your positive direction.
$\vec v_{\rm ac}$ is $30 \,\rm m\,s^{-1}\, East$ and $\vec v_{\rm bc}$ is $10\,\rm m\,s^{-1}\, West $.
$30\,\rm m\,s^{-1}\, East$ is the same as $-30\,\rm m\,s^{-1}\, West$.
$v_{\rm ac} =v_{\rm ab} + v_{\rm bc}\Rightarrow -30 = v_{\rm ab} + 10 \Rightarrow v_{\rm ab} = -40 $
So the velocity of $a$ relative to $b$ is $-40\,\rm m\,s^{-1}\, West$ is the same as $40\,\rm m\,s^{-1}\, East$
On the other hand Suppose that you chose East as your positive direction.
$\vec v_{\rm ac}$ is $30 \,\rm m\,s^{-1}\, East$ and $\vec v_{\rm bc}$ is $10\,\rm m\,s^{-1}\, West $.
$10\,\rm m\,s^{-1}\, West$ is the same as $-10\,\rm m\,s^{-1}\, East$.
$v_{\rm ac} =v_{\rm ab} + v_{\rm bc}\Rightarrow 30 = v_{\rm ab} + (-10) \Rightarrow v_{\rm ab} = 40 $
So the velocity of $a$ relative to $b$ is still $40\,\rm m\,s^{-1}\, East$
If it was a two or three dimensional problem then all you would need to do is to define two (or three) unit vectors (positive directions) and deal with each of the components in the the two (or three) directions as before.