You're looking to compute streamlines. You can compute them by computing the direction of the fluid flow at every point. At every $(x,y)$ coordinate at which you have a measurement, the vector of the fluid flow gives you the streamline.
In order to compute the streamline between grid points (assuming that the only information regarding the flow field that you have is the measurements at the discrete points), the simplest method is to linearly interpolate.
Assume you want a measurement at coordinate $(x_0+\epsilon, y_0+\delta)$, and you have measurements at $(x_0,y_0), (x_0,y_1), (x_1,y_0), (x_1,y_1)$. The most straightforward approach, which should be accurate enough for visualization, is to interpolate for values at $(x_0,y_0+\delta)$ and $(x_1,y_0+\delta)$. Then, use these results to interpolate for the final value.
Geometrically, imagine your four measurements are at the corners of a box. First, draw a line horizontally through where you want your estimate. Then, where this line intersects the vertical sides of the box, estimate the streamlines there. Then, use these estimates to estimate the value inside the box.
Linear interpolation can be performed using the relationship $ \frac{v_2-v_1}{u_2-u_1} = \frac{v-v_1}{u-u_1}. $ Here, the $v$'s represent your streamline measurements, and the $u$ represents each coordinate (either $x$ or $y$).
Example, to estimate the streamline at coordinate $(.4,.6)$: $\begin{align*} f(0,0) &= (3,4) \\ f(0,1) &= (3,5) \\ f(1,0) &= (2,4) \\ f(1,1) &= (2,2) \\ \end{align*} $
$ f(0,.6) = \left( \frac{3-3}{1-0}(.6-0)+3, \frac{5-4}{1-0}(.6-0)+4 \right) = (3,4.6) $ $ f(1,.6) = \left( \frac{2-2}{1-0}(.6-0)+2, \frac{2-4}{1-0}(.6-0)+4 \right) = (2,2.8) $
Now we've interpolated vertically (which is why we replaced $u$ in the general formula above with the $y$ coordinate). Next, we interpolate over $x$. $ f(.4,.6) = \left( \frac{2-3}{1-0}(.4-0)+2, \frac{2.8-4.6}{1-0}(.4-0)+4.6 \right) = (2.6,3.88) $