Operator sum representation
Quantum channels may be represented in an operator-sum form: For a channel $\Lambda$ acting on a $d$-dimensional state $\rho$, there exists some set of $d$-dimensional operators $\{E_1, \dots, E_k, \dots\}$ such that
\begin{align}
\Lambda(\rho) &= \sum_k E_k \rho E_k^\dagger \tag{1} \\
\sum_k E_k^\dagger E_k &= I_d, \tag{2}
\end{align}
where $I_d$ is the identity operator (see Sec. 8.2.3 of Nielsen and Chuang or Corollary 2.27 of Watrous' TQI). In particular, if you can express the channel using only Pauli operators then we can choose to define
$$
E_k = \sqrt{p_k}\sigma_k \tag{3},
$$
where we set $d=2$ for a single qubit and $p_k>0$ without loss of generality. In this case, tracing the LHS of Eq. (2) gives us
\begin{align}
\text{Tr}\left(\sum_k E_k^\dagger E_k\right) &= \sum_k p_k \text{Tr}\left(\sigma_k^\dagger \sigma_k \right) \tag{4a-c}
\\&= \sum_k p_k \text{Tr}\left( I_2\right) \\
&= 2\sum_k p_k,
\end{align}
where we have used $\sigma_k^2 = I$ for all four Pauli operators. Meanwhile tracing the RHS of Eq. (2) gives $\text{Tr}(I_2) = 2$. Then,
$$
\sum_{k=0}^3p_k = 1 \tag{5}
$$
is sufficient for $\Lambda$ to be a valid channel. So under the identification of Eq. (3), if you are given a channel that permits an operator sum form that uses only Paulis (or you choose to define a channel in that way), then the coefficients will form a probability distribution. From Lines (4a-c) this holds for an operator sum form written in terms of an orthonormal basis for $2\times 2$ Hermitian matrices.
Not all channels obey have coefficients summing to one
If you are given an operator sum representation consisting of operators other than Paulis, the above does not necessarily hold. For example, the "replace with $|0\rangle$" channel has operator sum representation
\begin{align}
E_0 &= |0\rangle \langle 0|\\
E_1 &= |0\rangle \langle 1|.
\end{align}
The corresponding channel sends every input $\rho$ to $|0\rangle$. Clearly the coefficients for each $E_k$ do not form a probability distribution. Indeed, using $ |0\rangle \langle 0|= \frac{1}{2}(I + \sigma_3)$ and $|0\rangle \langle 1| = \frac{1}{2} (\sigma_1 - i\sigma_2)$ you can verify that there's no way to write this channel in the form of Eq (1) that uses only Pauli operators, and so the above analysis doesn't hold.
