I will answer your question for a Galilean transformation because it extends naturally to the Lorentz transformation and it will be less confusing this way (I hope).
Consider two observers: observer A (blue) and observer B (red)
If we consider A's frame then B is moving to the right with velocity $\vec v=(v,0,0)$. The coordinates of B are related to A's by
\begin{align}
x'&=x-v\,t\tag{1}\\
t'&=t
\end{align}
Where B has the primed coordinates. Remember that the blue gridlines are all the points which have a constant $x$ or $t$ coordinate for observer A and the red gridlines have constant $x$ or $t$ coordinates for observer B. For example the red gridline in the middle has $x'=0$ (I didn't draw the time gridlines for observer B).
So to answer your question: the direction of $\vec v$ does matter. Before $t=0$ observer B is always moving towards A and after $t=0$ B is always moving away from A. The sign of $v$ can switch in which direction B is moving though. If $v<0$ then B is moving to the left. When you are considering left moving observers you have to be careful. You can define the left moving transformation in two ways. Either define $v$ to always be positive, so the transformation for B moving to the left becomes
\begin{align}
x'&=x+v\,t\tag{2}\\
t'&=t
\end{align}
or allow $v$ to be negative, which means you can still use equation (1).
The sign in the doppler shift matters because it matters if the emitter is either moving towards or away from you. Which one you need depends on the convention you use so make sure your answer makes sense (something moving towards you will increase the frequency).
Note that for the inverse transformation you always get an additional minus sign because A will be moving in the direction $-\vec v$ according to B's frame. So
\begin{align}
x&=x'+v\,t'\tag{3}\\
t&=t'
\end{align}
