Starting from the conservation of mass:
$$ \dot m_{1}=\dot m_{2} $$
This translates to
$$ \rho_{1} S_{1} V_{1}=\rho_{2} S_{2} V_{2} $$
Assuming incompressible flow, thus $\rho_{1}=\rho_{2}$
gives:
$$S_{1} V_{1}= S_{2} V_{2} $$
With $S_{1} V_{1} = Q_{1}$ , the formula you are using.
This formula follows directly from the mass balance, with only the assumption of incompressible flow. There is no assumption on turbulent or laminar flow, thus this equation holds for both flow types.
With the balance given above you can calculate the speed in at location 2 for the given value of $V_1$ and the ratio of Areas, just as you did. There is no need to account for the flow type.
However, it should be noted here that these are average speeds. If you want to go into further detail, you could include friction forces in the pipe. These friction forces depend on the flow type, and determine the shape in velocity profile, and the resulting velocity.
However, this is significantly more difficult to do than just solving a couple equations.
I know there are some rules of thumb to estimate losses in pipes, but you have to check if their assumptions are valid for your case.
Perhaps you can have a look at Pipe Flow Fluid Mechanics Course
It can be shown via simple dimensional analysis. We know that $[u]=m/s$, so just multiply by 1 in terms of an area:
$$
[u]=\frac{m}{s}\cdot\frac{m^2}{m^2}=\frac{m^3}{m^2\cdot s}=\color{red}{\frac{1}{m^2}}\cdot\color{blue}{\frac{m^3}{s}}
$$
The blue term is the volumetric flow rate while the red term is the area, thus we have a volumetric flow rate per unit area. Multiplying this volumetric flow rate by the mass density gives a final unit of
$$
\left[\rho u\right]=\color{red}{\frac1{m^2}}\cdot\color{blue}{\frac{kg}{s}}
$$
which is the mass flow rate (the amount in mass that flows through the surface).
Since flows are generally three-dimensional, we are interested in how much of a fluid (either in terms of the volume or the mass) goes through an arbitrary surface (area) in some unit of time.
Best Answer
The volume of a differential fluid element (or differential anything, really) is the area of its cross section times the perpendicular distance over which that cross section is stretched.
In this picture, that'd be $A\cdot h$, not $A\cdot b$, since the distance $b$ is not perpendicular to the cross section.
Now, if you want to know how much fluid passes through an area element per unit time, you're essentially calculating the volume of fluid that passes through that area in one time unit. Look at the above picture: say we want to calculate the fluid flow through the bottom face, and say the velocity $\vec{v}$ points in the direction of side b, so that it's not perpendicular to the surface. Then the perpendicular height (needed to calculate volume) of this fluid element comes only from the part of the velocity parallel to the normal, $\hat{n}$ (which is pointed straight up). The perpendicular height would be the vertical component of velocity $\vec{v} \cdot \hat{n}$ time the time, $t$. Multiply by the area A (and divide by $t$ to get the fluid flow per unit time), and you get
$$ \vec{v} \cdot \vec{dA}$$
where $\vec{dA}$ is $A$ times the unit surface normal, which is the usual formula.