In the general case of a representation $(j_-, j_+)$ of Lorentz, chirality is defined as $j_+ - j_-$. The chiralities $\pm s$ for $(s,0)$ and $(0,s)$ are a particular case of this definition.
So then I assume the answer to your first question is "by definition", and the next question is, "How can this definition of chirality be related to helicity ?".
To find a relation between chirality and helicity, let's consider the wave function
$$ \langle 0 | O_{(m_-,m_+)}(0) | \bar{p}, \sigma\rangle $$
where $O_{(m_-, m_+)}$, with $m_- = - j_-, ..., j_-, m_+ = -j_+, ..., j_+$ denotes an operator transforming in the $(j_-, j_+)$ representation of Lorentz, and $| \bar{p}, \sigma \rangle$ is a massless state in the reference $\bar{p}^\mu=(p,0,0,p)$ with helicity $\sigma$.
We can show, by group theoretic arguments (proof below), that this matrix element is non-vanishing only in the case where $m_- + m_+ = - \sigma$ and $j_+ - j_- = \sigma$. The second relation establishes the relation between chirality and helicity. More precisely, the statement is that massless particles of helicity $\sigma$ can only be interpolated by operators of chirality $j_+ - j_- = \sigma$.
Proof of the statement
Defining the Pauli-Lebanski vector $ W^\mu = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} J_{\nu \rho} P_\sigma $, and denoting $J_\pm^i$ the generators in the $j_\pm$ $SU(2)$ irreps respectively, we have the following relations / definitions :
$$ \begin{cases} W^{\pm} \equiv W^1 \pm i W^2 \\ J_{\pm}^i = \frac{1}{2} (J^i \pm i K^i) \\ W^1 = K^2 - J^1 \\ W^2 = - K^1 - J^2 \end{cases} $$
After some easy manipulations, we find
$$ W^+ = - 2 \sqrt{2} J_+^+ \\ W^- = - 2 \sqrt{2} J_-^- $$
where, $J^+_+$ is the raising operator in the $j_+$ rep, and $J_-^-$ is the lowering operator in the $j_-$ rep.
Since in the massless case, the $W^\pm$ operators should be 0, we have
$$ J_-^- | \bar{p}, \sigma \rangle = J_+^+ | \bar{p}, \sigma \rangle = 0 $$
Remember the action of the $J_-^-$, $J_+^+$, $J^3$ operators :
$$ [J^3,O_{(m_-,m_+)}] = [J_-^3, O_{(m_-,m_+)}] + [J_+^3,O_{(m_-,m_+)}] = (m_- + m_+) O_{(m_-,m_+)} \\ [J_-^-,O_{(m_-,m_+)}] \propto O_{(m_- - 1,m_+)} \\ [J_+^+,O_{(m_-,m_+)}] \propto O_{(m_-,m_+ + 1)} $$
First note that the first relation implies
$$ - \sigma \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle = \langle 0 | [J^3,O_{(m_-, m_+)}] (0) | \bar{p}, \sigma \rangle = (m_- + m_+) \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle $$
which implies $-\sigma = m_- + m_+$ for the matrix element to be non-zero.
Now, assume that $j_- > m_-$, so that that there is a state $O_{m_- + 1, m_+}$. In this case, the second equation implies
$$ 0 = \langle 0 | [J_-^-, O_{m_- + 1, m_+}] | \bar{p}, \sigma \rangle \propto \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle $$
The matrix element would be vanishing. For it not to vanish, we thus need $m_- = j_-$ to be the highest state.
Similarly, assume $m_+> - j_+$ so that there is an operator $O_{m_-, m_+ - 1}$. Then
$$ 0 = \langle 0 | [J_+^+, O_{m_- , m_+ - 1}] | \bar{p}, \sigma \rangle \propto \langle 0 | O_{(m_-, m_+)} (0) | \bar{p}, \sigma \rangle $$
Thus for the matrix element not to vanish, $m_+ = - j_+$.
Comining these conditions with the first one
$$ j_+ - j_- = - m_+ - m_- = \sigma $$
Otherwise, the matrix element vanishes.
