I'm taking the Clifford algebra route as pointed out by non-user38741 and Giorgio Comitini, but I'll try to justify intuitively how to end up there and how the spinor transformation law appears inevitable. So I start with geometric algebra, which is simply another name for Clifford algebra when used in a physics context, and the vectors are taken to be elements of the algebra themselves (i.e. we're not imposing a separate matrix algebra). So take $\mathbb{R}^{n, m}$ with inner product $<\cdot,\cdot>$, and define the geometric algebra $\mathcal{G}(\mathbb{R}^{n, m})$ as the freest associative algebra over $\mathbb{R}^{n, m}$ which satisfies
\begin{equation}
v^2 = <v, v>,
\end{equation}
where the square is, of course, the algebra multiplication. We will call the multiplication in this algebra the geometric product.
Admittedly, this does introduce another space, but that is an extremely natural one: the elements of the geometric algebra can be interpreted to consist of the scalars, the vectors of $\mathbb{R}^{n, m}$, the bivectors $u\wedge v$ where $u$ and $v$ are vectors and $u\wedge v := \frac{1}{2}(uv - vu)$, the 3-vectors $u\wedge v\wedge w$ and so on, up to (n + m) -vectors. The $n$ -vectors can be interpreted as directed area/volume/n-volume elements. For a whimsical introduction, see "Imaginary numbers are not real", or as a thorough introduction either Hestenes' "Clifford algebra to Geometric Calculus" or Doran and Lasenby's Geometric Algebra for Physicists.
Now, it turns out that a rotation of the vector $v$ in the plane defined by a simple bivector $\omega$ by $|\omega|$ radians (where the absolute value is $\sqrt{-\omega^2}$, since the square of $\omega$ is negative) can be expressed in geometric algebra (GA) as
\begin{equation}
v \mapsto \exp(\omega) v \exp(-\omega),
\end{equation}
where the exponential is defined by the usual power series, with the multiplication being the geometric product, and a simple bivector is a bivector that can be written as the wedge product $a \wedge b$ for some vectors $a, b$. A general rotation is then given by the same formula, but with the $\omega$ not being necessarily simple (i.e. it may need to be a sum of several simple bivectors). The result of the exponential is then in the even subalgebra, i.e. built out of objects that can be expressed as a sum of products of an even number of vector factors. We call the result of the exponentiation a rotor, and often denote $R = \exp(\omega)$. Then the object on the right hand side of the transformation can also be written as $\tilde{R}$, where the tilde denotes reversion, which simply means taking each factor in a geometric product and reversing their order. Further, $R \tilde{R} = 1$ when $R$ is a rotor.
The first glimpse of a spinor-like transformation law appears: in general, we can rotate all elements of the space by the two-sided rotation law given above, and nothing changes. However, if we represent rotations by the rotor $\exp(\omega)$, then the composition of rotations is given by $\exp(\omega_1) \exp(\omega_2)$, which is also a rotor.
Now, let us stick specifically to $\mathbb{R}^{1, 3}$. Then we can write the free Dirac equation as
\begin{equation}
\nabla \psi I_3 + m \psi = 0,
\end{equation}
where $\nabla$ is the vector derivative $\nabla = e^\mu \partial_\mu$, and the $e^\mu$ are basis vectors acting via the geometric product (so that $\nabla$ itself is algebraically a vector). The Dirac field $\psi$ takes values in the even subalgebra of the geometric algebra. $I_3$ is a three-vector which appears to pick a preferred slice of spacetime, and therefore break Lorenz invariance. However, consider another choice given by $I'_3 = R I_3 \tilde{R}$. Then the corresponding new Dirac equation is
\begin{equation}
\nabla \psi' R I_3 \tilde{R}+ m \psi' = 0.
\end{equation}
Now if $\psi$ solves the original Dirac equation, then clearly $\psi' = \psi \tilde{R}$ solves this new equation with $I_3'$. In other words, when the object $I_3$ transforms like a (three)-vector under rotations, then $\psi$ transforms like a spinor, and the transformation law has appeared.
Then note that the physical predictions of the theory only depend on the Dirac bilinears, which in this language can be written analogously to
\begin{equation}
\psi I_3 \tilde{\psi},
\end{equation}
and that when $I_3$ transforms as a three-vector and $\psi$ as a spinor, the physical predictions remain unchanged. In other words, the spinor transformation law is required here to keep the physical predictions of the theory independent of the choice of the directed volume element $I_3$.
Indeed, there is a natural interpretation of the object $\psi$ as a product of a rotor, scaling and a transform between scalars and pseudoscalars in $\mathbb{R}^{1,3}$. In this way, the spinor transformation law appears naturally as the composition of rotors (or rotor-like objects). Of course, since there is no treatment of quantum field theory in the geometric algebra language, it's not clear how far or seriously this can be taken as an interpretation of the physical Dirac equation, but nonetheless it at least provides an example where spinors appear naturally, without manually imposing the transformation law. Rather, it comes from transformations of the solutions of the Dirac equation when the choice of the constant $I_3$ transforms by rotations.
I'm sure that this flash-introduction to the subject leaves many questions unanswered and it may be a bit confusing, but if I piqued your interest I suggest you follow some of the links here and continue further that way.
