Intuition behind field transfomations Consider a real field $V^{\mu}(x)$ defined on a 4-dimensional Minkowski space. Acted by a transformation $\Lambda = \Lambda^{\mu}{}_{\nu} $ it transforms like
$$V^{\mu}(x) \to V^{'\mu}(x) = \Lambda^{\mu}{}_{\nu}  V^{\nu}(\Lambda^{-1}x)$$
My question is: what is the intuition behind this transformation? I can't wrap my head around it.
 A: Minor general advice. If you want to obtain intuition for any concept, it is smart to turn back to scenarios that you can easily picture in your head, and reflect your experience in those toy models to the present idea that you are trying to grasp.
Now returning to your question. That equation gives us the recipe of how a vector field transforms under a Lorentz transformation. Now Lorentz transformations are the linear transformations that leave $x ^T \eta x$ invariant where $x$ is 4-vector in Minkowski space and $\eta$ is the metric. A transformation that leaves the inner product invariant for short.
To really feel and understand the physics behind your question, let us think of another transformation that leaves the inner product invariant in another space than Minkowski space to, preforably something we can imagine a lot easier. For example the Euclidean plane!
The linear transformations that leave the inner product of the Euclidean plane are rotations. Consider a vector field $F$ on $\mathbb{R} ^2$. Let's see how this vector field transforms when we apply a rotation. When you apply a rotation, you rotate  the points of the plane, that's obvious. But most importantly you also rotate the vectors that sit on each point of this plane. So when you apply a rotation, you take all the points on the plane and rotate them by the angle of rotation, but you must also rotate the vectors that sit on those points by the same angle.
As you are probably familiar with, when we are dealing with scalar fields, we only rotate the points of the space, not the fields itself because there is nothing to rotate. Scalar fields are just numbers sitting on each point.
That is the whole spirit of the transformation law you have provided.
A: The components of the vector field transforms the same way as the components of the position vector.
For a scalar field with the value of 7 at point (3,0), for another reference frame rotated $90 ^\circ$, the rotated point (0,-3) has the same value.
But for a vectorial field with value (0,1) at point (3,0), for the new frame, the rotated point (0,-3) has now the value (1,0).
The same rotation matrix
$\begin {bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$
is applied to the vector position and to the vector field value itself.
It is easy to draw a picture and see what happens in this toy model 2-D.
