Representation of $SU(2)$, i.e., spin 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?
 A: 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.  
A: 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.
--
$^1$ Hermitian or anti-Hermitian depending on conventions for factors of $i$.
