Passive Lorentz transformations are what everybody learns first. There is one physical reality, and you're just describing that one physical reality using two different coordinate systems, where one coordinate system uses a set of axes that are rotated and/or boosted relative to the axes used by the other coordinate system. You're just using a different set of numbers to describe every event in the universe (ignoring the complications due to gravity). It makes sense to use notation of $x'$ vs $x$ with a passive Lorentz transformation, because the prime is denoting coordinates as defined in two different coordinate systems.
With an active Lorentz transformation, you're contemplating two different physical realities, where some piece of that physical reality has been rotated and/or boosted relative to how it is in the other physical reality. If the part of physical reality you're concerned with is your TV set, the relationship between your TV set in its normal orientation vs your TV set after you've physically picked it up and rotated it 90 degrees clockwise from its normal orientation would be described by an active Lorentz transformation. You don't use two different coordinate systems to describe the two different TV orientations you're considering; you just use the one lab frame to describe the TV set in both of its orientations. So using an $x'$ doesn't make sense, because you aren't using a primed coordinate system for anything. Saying
$$\phi (x) \rightarrow \phi'(x) = \phi \left(\Lambda^{-1} x \right)$$
makes sense as is, with no prime on any of the $x$'s, because you're talking about two physically different fields, both described in the same coordinate system. And the expression
$$y=\Lambda^{-1}x$$
makes sense as is, because $y$ is a different point within the same coordinate system as $x$, and you most commonly label different points within one coordinate system by using different letters, not by introducing primes. Using primes would risk confusing a reader into erroneously thinking that some primed coordinate system was being used, in addition to the lab frame. Similarly, it makes sense to say
$$(\partial_\mu \phi')(x) = \left( \Lambda^{-1} \right)^\nu_{\phantom{\nu} \mu} (\partial_\nu \phi)(y)\ ,$$
because it's two physically different fields that are being considered, both of which are described using the same coordinate system.
A passive transformation is a mere change of coordinates on a manifold and there is usually nothing special about this. Every tensor equation is trivially invariant (or better to say covariant) under such transformations. In terms of manifolds and atlases, this corresponds to just a change of charts. So let $M$ be the said manifold, say e.g. real, differentiable and of dimension $n$, and let $(U,\phi)$ and $(V,\psi)$ be two charts on $M$ such that $U\cap V\neq\varnothing$. The map $\psi\circ\phi^{-1}:\phi(U\cap V)\to\psi(U\cap V)$ is a diffeomorphism that represents a change of coordinates on the overlap $U\cap V$ over $M$ and involves open subsets of $\mathbb R^n$.
An active transformation is an actual motion on the manifold $M$ through a diffeomorphism $\phi:M\to M$ which sends any point $p$ to the new point $q = \phi(p)$. When performing an active transformation, not only you move the point $p$ to $q$, but also every tensor/form on $p$ is to be moved from the tensor/exterior algebra $T(T_pM)$/$\bigwedge^*(T^*_pM)$ to $T(T_qM)$/$\bigwedge^*(T^*_qM)$ through the maps $\phi(p)_*$ and $\phi(p)^*$ induced by $\phi$. An interesting case is when $\phi$ arises from the flow of a vector field (e.g. a Killing vector field). Then the notion of symmetry is studied by means of the Lie derivative of a tensor by the vector field. Thus, the description in terms of local coordinates is the following. Fix a point $p\in M$ and charts $(U,\rho)$, $(V,\psi)$ such that $p\in U$ and $\phi(p)\in V$. The map $\psi\circ\phi\circ\rho^{-1}:\rho(U\cap\phi^{-1}(V))\to\psi(\phi(U)\cap V)$ is a diffeomorphism between open subsets of $\mathbb R^n$, and its differential relates tensor quantities between tangent spaces.
Best Answer
Consider $x^{'\mu} = x^\mu - \epsilon^\mu$. The active transformation of fields would require the change of $\phi(x)$ to $\phi'(x)$. The new field at x is given by $\phi'(x) = \phi(x) + \epsilon^\mu \partial_\mu \phi(x)$.
Whereas if the transformation was passive the fields would be written as :
$\phi(x') = \phi(x) - \epsilon^\mu \partial_\mu \phi(x)$.
The only difference is the position of $'$ in both the cases. I hope this answers your questions.