"Twistors are spinors of the compactified Minkowski space" is not quite true. Twistors in Minkowski space are spinor fields which satisfy a particular PDE: the twistor spinor equation. Relative to flat coordinates and the corresponding global frame for the tangent bundle, a twistor spinor $\psi$ is given in terms of a pair of constant spinors $(\varphi,\eta)$ by $$ \psi(x) = \varphi + x \cdot \eta $$ where $\cdot$ is the Clifford action. Since the twistor spinor equation is conformally invariant, one can extend this to a spinor on teh conformal compactification of Minkowski space.

The situation in general is the following.

Let $(M,g)$ be a pseudoriemannian spin manifold. Let $\Sigma \to M$ be a bundle of modules of the Clifford bundle $Cl(TM)$. The spin connection defines a map on sections $$ \nabla : C^\infty(M;\Sigma) \to \Omega^1(M;\Sigma)~. $$ The Clifford action of 1-forms on spinor fields defines a bundle map $$ c: T^* M \otimes \Sigma \to \Sigma $$ which induces a map on sections $$ \Omega^1(M;\Sigma) \to C^\infty(M;\Sigma)~. $$ Composing this map with the covariant derivative above defines a differential operator $$ D : C^\infty(M;\Sigma) \to C^\infty(M;\Sigma) $$ This is the Dirac operator.

Now the kernel of $c : T^* M \otimes \Sigma \to \Sigma $ defines a sub-bundle $W subset T^* M \otimes \Sigma$, so that $$ T^*M \otimes \Sigma \cong W \oplus \Sigma~.$$ Composing the above covariant derivative with the projection onto $W$ along $\Sigma$ defines a differential operator $$ P : C^\infty(M;\Sigma) \to C^\infty(M;W) $$ called the Penrose operator, whose kernel are called twistors.

In other words, a spinor field $\psi \in C^\infty(M;\Sigma)$ is a twistor (spinor) if $P\psi = 0$. We can write this equation more explicitly as follows: $$P_X \psi = \nabla_X \psi + \frac1n X \cdot D\psi~,$$ where $X \in C^\infty(M;TM)$, $\dim M = n$ and I have used the original Clifford conventions for the Clifford product, so that there is a minus sign in $$ X\cdot X \cdot \psi = - g(X,X) \psi~.$$

It is an instructive exercise to work out that when $(M,g)$ is $4$-dimensional Minkowski spacetime, the solutions to the twistor spinor equation are precisely given by the first formula in this answer.

"Twistors are spinors of the compactified Minkowski space" is not quite true. Twistors in Minkowski space are spinor fields which satisfy a particular PDE: the twistor spinor equation. Relative to flat coordinates and the corresponding global frame for the tangent bundle, a twistor spinor $\psi$ is given in terms of a pair of constant spinors $(\varphi,\eta)$ by $$ \psi(x) = \varphi + x \cdot \eta $$ where $\cdot$ is the Clifford action. Since the twistor spinor equation is conformally invariant, one can extend this to a spinor field on the conformal compactification of Minkowski space.

The situation in general is the following.

Let $(M,g)$ be a pseudoriemannian spin manifold. Let $\Sigma \to M$ be a bundle of modules of the Clifford bundle $Cl(TM)$. The spin connection defines a map on sections $$ \nabla : C^\infty(M;\Sigma) \to \Omega^1(M;\Sigma)~. $$ The Clifford action of 1-forms on spinor fields defines a bundle map $$ c: T^* M \otimes \Sigma \to \Sigma $$ which induces a map on sections $$ \Omega^1(M;\Sigma) \to C^\infty(M;\Sigma)~. $$ Composing this map with the covariant derivative above defines a differential operator $$ D : C^\infty(M;\Sigma) \to C^\infty(M;\Sigma) $$ This is the Dirac operator.

Now the kernel of $c : T^* M \otimes \Sigma \to \Sigma $ defines a sub-bundle $W \subset T^* M \otimes \Sigma$, so that $$ T^*M \otimes \Sigma \cong W \oplus \Sigma~.$$ Composing the above covariant derivative with the projection onto $W$ along $\Sigma$ defines a differential operator $$ P : C^\infty(M;\Sigma) \to C^\infty(M;W) $$ called the Penrose operator, whose kernel are called twistors.

In other words, a spinor field $\psi \in C^\infty(M;\Sigma)$ is a twistor (spinor) if $P\psi = 0$. We can write this equation more explicitly as follows: $$P_X \psi = \nabla_X \psi + \frac1n X \cdot D\psi~,$$ where $X \in C^\infty(M;TM)$, $\dim M = n$ and I have used the original Clifford conventions for the Clifford product, so that there is a minus sign in $$ X\cdot X \cdot \psi = - g(X,X) \psi~.$$

It is an instructive exercise to work out that when $(M,g)$ is $4$-dimensional Minkowski spacetime, the solutions to the twistor spinor equation are precisely given by the first formula in this answer.