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Let \begin{equation} X= \begin{bmatrix} 0 & 1\\ 0 & 0\\ \end{bmatrix}, \qquad Y= \begin{bmatrix} 0 & 0\\ 1 & 0\\ \end{bmatrix}, \qquad H= \begin{bmatrix} 1 & 0\\ 0 & -1\\ \end{bmatrix}\tag{1} \end{equation} If $V_m$ is the $(m+1)$-dimensional complex representation of $\text{sl}(2,\mathbb{C})$, then we know that there exists a basis $u_m,u_{m-2},...,u_{-m}$ such that $u_k$ is the eigenvector of $H$ with eigenvalue $k$ and $Y u_k = u_{k-2}$.

In physics, on the other hand, we write $S_{\pm} =X,Y$ and $S_z = H/2$ and use the basis $|s,m_s \rangle$ such that $$\begin{align} S_z |s,m_s \rangle &= m_s |s,m_s \rangle \\ S_{\pm} |s,m_s \rangle &= \sqrt{s(s+1)-m_s (m_s \pm 1)} |s,m_s \pm 1\rangle \end{align}\tag{2}$$ and that $|s,m_s \rangle$ are orthogonal.

Is there a deeper reason why we put the coefficients in front of $|s,m_s \rangle$, when $S_{\pm}$ acts on it and why $|s,m_s \rangle$ are orthogonal?

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In the specific case of the 2-dimensional representation, the coefficients are 1 so it doesn't matter much. On the other hand, for the higher-dimensional reps of $SU(2)$, the coefficients in front are not trivial, v.g. your raising operator $$ X\to \sqrt{2}\left(\begin{array}{ccc}0&1&0\\0&0&1\\ 0&0&0\end{array}\right) $$ and for even larger representations the coefficients are not all the same, $v.g.$ $$ X\to \left(\begin{array}{ccc}0&2&0&0&0\\0&0&\sqrt{6}&0&0\\ 0&0&0&\sqrt{6}&0\\ 0&0&0&0&2\end{array}\right)\, . $$

If you don't have the correct coefficients your matrices will not be a hermitian representation and thus will not exponentiate to a unitary rep. Alternatively, if your basis states are not properly normalized you will not get a unitary rep. either. We want unitary because it preserves the (complex) inner product $\langle \phi\vert\psi\rangle$ and (for instance) probabilities of outcomes depend on such overlaps.

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The difference between an abstract (finite-dimensional) $su(2)$ Lie algebra representation $\rho:su(2)\to V$, and applications in quantum physics, is that the vector space $V$ in quantum physics is typically a complex Hilbert space. In other words, in quantum physics $V$ additionally comes equipped with a sesquilinear form, and the image $\rho(su(2))$ should consist of (anti)Hermitian$^1$ operators, i.e. the representation should be unitary. This in turn guarantees that operators are diagonalizable (although not simultaneously), and eigenvectors are orthogonal. The eigenvalues in OP's eq. (2) are (to a large extent) dictated by the unitary representation.

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$^1$ Hermitian or anti-Hermitian depending on conventions for factors of $i$.

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