Like many of the other questions I answer on Earth Science SE, the answer is complicated. Based on the principle of geostrophy, the Coriolis force will point to the right of the wind vector in the northern hemisphere (left in the southern hemisphere).
Can you validate the Coriolis force based on the wind rose? No. You would need multiple simultaneous wind observations over a wide observation space. You can see this in weather maps that show wind direction. It becomes easier to see the Coriolis effect on upper air maps, as they aren't affect by the surface. It is also easier to see further from the equator, where the Coriolis is zero. Earth NullSchool has nifty maps that show worldwide winds that can help you see the Coriolis effect (the color is the mean sea level pressure):
In your wind rose, you can approximate the Coriolis effect though, by turning it sideways.
The Coriolis effect can be experienced beyond the realm of meteorology (and I am not referring to the toilet-bowl fallacy). Long before Gaspard Coriolis lived, it was recognized that cannonballs were deflected by the Earth's rotation, though it did have some inaccuracies.
One question I anticipate that I anticipate you might have is "How far would a something (wind or a cannonball) have to travel before the Coriolis effect would have a meaningful effect?" Projectiles are, in my opinion, the easiest way to show that this is a real thing. This "acceleration" is approximately $a_{cor}=fv$ where $f$ is the latitudinally dependent Coriolis parameter. and $v$ is the speed. You can see how this impacts ballistics through this Wikipedia entry. Given various assumptions, like a constant latitude, ignoring drag, and ignoring the loss of forward momentum from the perpendicular acceleration, the distance one would be off after firing a projectile would be: $$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u_0^{-1} $$, where $\Omega$ is the angular velocity of the earth, $\phi$ is the latitude, and $u$ is the speed, and $\Delta x$ is the distance to the target. So at say 45 degrees north and going 5 meters per second over a 100 meter distance, you would be off by 0.2 meters, but if you are the Paris gun with a horizontal velocity of 1009.7 m/s, trying to bombard the city 120 km away at a latitude of 49 degrees north, you'll be 1.3 km off if you don't account for the Coriolis effect.
And because I know I will get someone who wants me to show my work for the previous equation here it is: $$\frac{d\vec{v}}{dt}=-f\hat{k}\times\vec{v}$$. If we consider the initial conditions $\vec{v}(t=0)=u(t=0)\hat{i}+v(t=0)\hat{j}=u_0\hat{i}+0\hat{j}$, then we have the system $$\frac{du}{dt}=fv$$ $$\frac{dv}{dt}=-fu$$. Since I made the assumption that momentum along the trajectory is not altered by the Coriolis force, the system reduces to $$\frac{du}{dt}=0$$ $$\frac{dv}{dt}=-fu$$. If we think that wind is the changing coordinates ($u=\frac{dx}{dt}$, $v=\frac{dy}{dt}$), then the system can be represented as $$\frac{d^2x}{dt^2}=0$$ $$\frac{d^2y}{dt^2}=-fu=-fu_0$$. Therefore $$\Delta y= fu_0t^2$$. And since the projectile must travel distance $\Delta x$ at velocity $u_0$, then $$t=\frac{\Delta x}{u_0}$$ which makes $$\Delta y = 2 \Omega \sin(\phi) (\Delta x)^2 u_0^{-1} $$
