Can a spinor be defined as any quantity which transforms linearly under Lorentz transformations?

In many contexts, we would like to determine how Lorentz transformations act on the mathematical objects that characterize a particular theory. In the case of classical, Lorentz-invariant field theories on Minkowski space for example, we need to specify how Lorentz transformations act on the fields of the theory. This leads naturally to determining how Lorentz transformations can act on both Minkowski space, and the target spaces of the fields. This, in turn, leads naturally to the notion of a finite-dimensional representation of the Lorentz group.

On the other hand, in quantum mechanics, and by extension in quantum field theory, we often want to specify how Lorentz transformations act on the Hilbert space of the theory. In this context, Wigner's theorem in symmetries in quantum mechanics demands that up to phase, Lorentz transformations as unitary or anti-unitary operators on the Hilbert space. In turn, the fact that these transformations are only defined up to phase implies that one, in general, needs to consider the projective representations of the Lorentz group in addition to its "ordinary" representations.

Now, it turns out that determining the projective representations of the Lorentz group $\mathrm{SO}(3,1)^+$ is equivalent to determining the ordinary representations of its universal cover, which is called $\mathrm{Spin}(3,1)$! It's a spin group! In fact, for any $p,q$, the group $\mathrm{SO}(p,q)^+$ of isometries of the space $\mathbb R^{p,q}$ has universal cover $\mathrm{Spin}(p,q)$.

Therefore,

determining all projective representations of the Lorentz group is equivalent to determining the ordinary representations of the corresponding spin group.

It is overwhelmingly likely, in my opinion, that this is what the authors are referring to in their quote because the "objects that transform linearly under Lorentz transformations" that we consider in physics are precisely those objects that transform under projective representations of the Lorentz group (ordinary representations are included as a subclass), and these are precisely those objects that transform under ordinary representations of spin groups, and such objects are called spinors.

By the way, you'll probably find the following related post illuminating.

https://physics.stackexchange.com/a/96060/19976

Old, incomplete answer.

It's hard to know the intentions of the authors for certain, but here's some information that might help interpret what they're saying.

Recall that every representation of the Lorentz algebra $\mathfrak{so}(3,1)$ can be constructed from representations of $\mathrm{sl}(2,\mathbb C)$, the complexified angular momentum algebra (which is of course the algebra whose representations describe spin). The standard way of doing this is to note that if one complexifies the Lorentz algebra, when one finds that the complexification yields a direct sum of $\mathfrak{sl}(2,\mathbb C)$ with itself; \begin{align} \mathfrak{so}(3,1)_\mathbb C \approx \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C). \end{align} It follows that the representation theory of the Lorentz algebra reduces to the representation theory of the angular momentum algebra. In fact, every irreducible representation of the Lorentz algebra is essentially a tensor product of two irreducible representation of the angular momentum algebra, and these representations are often labeled by a pair $(s_1, s_1)$ of "spins" $s_1, s_1\in\{0,\frac{1}{2}, 1, \dots\}$. For example, the $(\frac{1}{2},0)$ representation is called the left-handed Weyl spinor representation, and the $(0,\frac{1}{2})$ is called the right-handed Weyl spinor representation.

The vector representation that you refer to, namely that standard representation that transforms a four-vector by a Lorentz transformation $\Lambda\in \mathrm{SO}(3,1)^+$ by mapping $V^\mu$ to $\Lambda^\mu_{\phantom\mu\nu} V^\nu$, corresponds to the $(\frac{1}{2}, \frac{1}{2})$ representation. So it is in a sense true as you say that "a 4-vector is capable of being written in spinor notation."

It's also the case that any finite-dimensional representation of the Lorentz algebra can be written as a direct sum of the irreducible representations $(s_1, s_2)$, so in a very real sense, all of the finite-dimensional objects that Lorentz-transform can be "built" out of "spin" representations, namely representations of the angular momentum algebra.

This answer is coming several years late, but I think the intended meaning is actually quite simple. Start with the transformation of a Lorentz vector, $$V^\mu \to \Lambda^\mu_\nu V^\nu.$$ Then the transformation law of a rank $2$ tensor can be built by product, $$T^{\mu_1\mu_2} \to \Lambda^{\mu_1}_{\nu_1} \Lambda^{\mu_2}_{\nu_2} T^{\nu_1 \nu_2}$$ with similar expressions for tensors of arbitrary rank. Clearly, this transformation is perfectly linear in $T$, as it must be by the definition of a representation.

However, if you regard the matrix $\Lambda$ as a fundamental thing that specifies a Lorentz transformation, then you might say that the transformation of a rank $2$ tensor is quadratic, i.e. quadratic in $\Lambda$. More precisely, all the tensor representations can be built up by tensor products of the vector representation, so they are "higher order".

Now the quoted statement is just taking this one level further. When we consider projective representations of the Lorentz group, there are even smaller, more "elementary" representations which are the right-handed and left-handed Weyl spinors, which transform as $$\psi_\alpha \to M^\beta_\alpha \psi_\beta, \quad \bar{\chi}_{\dot{\alpha}} \to (M^*)^{\dot{\beta}}_{\dot{\alpha}} \,\bar{\chi}_{\dot{\beta}}.$$ Just as the rank $2$ tensor can be built from tensoring two Lorentz vectors, the Lorentz vector can be built by tensoring two spinors. Therefore we might say that the spinors are "linear" and the vector is "quadratic" (in $M$). Indeed you can prove that all projective representations of the Lorentz group can be found by tensoring the spinors.

The real issue here is the language barrier. It's not that people are blindly parroting each other, any more than you are parroting your textbook when you say "all representations are linear". It's just that they are using a word in a different way than you expect.