$(B.1)$ is the functional equation of the Hurwitz zeta function and Knopp and Robins' proof is available here.
$$\tag{B.1}\zeta(z,a)=\frac{2\,\Gamma(1-z)}{(2\pi)^{1-z}}\left[\sin\frac {z\pi}2\sum_{n=1}^\infty\frac{\cos2\pi an}{n^{1-z}}+\cos\frac {z\pi}2\sum_{n=1}^\infty\frac{\sin2\pi an}{n^{1-z}}\right]$$
The author of your paper proposed a derivation of $(B.3)$ in reverse order since starting from $B.7)$. But let's try another derivation using the logarithm of the infinite products with $x:=\pi\,q\,|a|$ :
$$\tag{1}\sinh(x)=x\prod_{k=1}^\infty \left(1+\frac {x^2}{\pi^2k^2}\right)$$
$$\tag{2}\cosh(x)=\prod_{k=1}^\infty \left(1+\frac {4\,x^2}{\pi^2(2k-1)^2}\right)$$
which may be found for example here or in the online references.
The derivation will not be direct since these products are convergent while $(B.3)$ is clearly divergent and needs some regularization.
\begin{align}
\frac 12\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)&=\frac 12\log\bigl(a^2\bigr)+\sum_{n=1}^\infty\log\left(\frac {n^2}{q^2}+a^2\right)\\
&=\log|a|+\sum_{n=1}^\infty\log\frac {n^2}{q^2}+\log\left(1+\frac{q^2a^2}{n^2}\right)\\
\tag{3}&=\log|a|+\sum_{n=1}^\infty\log\left(1+\frac{q^2a^2}{n^2}\right)+2\sum_{n=1}^\infty\log n-\log q\\
\end{align}
The last sum at the right (as often in QFT) is heavily divergent so let's use zeta regularization to rewrite it in a finite form :
$$f(z):=\sum_{n=1}^\infty\frac{\log n-\log q}{n^z}=-\zeta'(z)-\zeta(z)\log q,\quad \text{for}\ \Re(z)>1$$
(since $\;\displaystyle\frac d{dz}n^{-z}=\frac d{dz}e^{-z\log n}=-\frac{\log n}{n^z}$)
From this we deduce the 'zeta regularized sum' (using analytic extension of $f(z)$ down to $0$) :
$$\sum_{n=1}^\infty\log n-\log q=\lim_{z\to 0^+}f(z)=-\zeta'(0)-\zeta(0)\log q=\frac{\log 2\pi}2+\frac 12\log q=\frac{\log 2\pi q}2$$
and get :
\begin{align}
\frac 12\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)&=\log|a|+\sum_{n=1}^\infty\log\left(1+\frac{q^2a^2}{n^2}\right)+\log 2\pi +\log q\\
&=\log\left[2\,\pi q|a|\prod_{n=1}^\infty \left(1+\frac{q^2a^2}{n^2}\right)\right]\\
\\
&=\log[2\,\sinh(\pi\;q\,|a|)],\quad\text{using}\ (1)\ \text{for}\;\;q\,|a|=\frac x{\pi}\\
\end{align}
Which is the first part of $(B.3)$ :
$$\tag{B.3}\sum_{n\in\mathbb{Z}}\log\left(\frac {n^2}{q^2}+a^2\right)=2\;\log[2\,\sinh(\pi\;q\,|a|)]$$
I'll let you get the corresponding equation for $\cosh$.
Hoping this clarified things.
I think that you misread the equation. In the paper, the rhs is $\frac{\log (x)}{m}$ while the expansion at the top of page $85$ is in terms of $\frac m{\log (x)}$.
Anyway, considering
$$-\frac{\zeta '(x)}{\zeta (x)-1}=k$$ around $x=1$, we have
$$\zeta (x)=\frac{1}{x-1}+\gamma -\gamma _1 (x-1)+\frac{1}{2} \gamma _2 (x-1)^2+O\left((x-1)^3\right)$$ So, we know the expansion of $\zeta '(x)$ and lon division gives
$$-\frac{\zeta '(x)}{\zeta (x)-1}=\frac{1}{x-1}+(1-\gamma )+$$ $$\left(2 \gamma _1+1-2 \gamma +\gamma ^2\right)
(x-1)+$$ $$\left(2 \gamma _1+(1-\gamma ) \gamma _1-2 \gamma \gamma _1-\frac{3 \gamma
_2}{2}+1-3 \gamma +3 \gamma ^2-\gamma ^3\right) (x-1)^2+O\left((x-1)^3\right)$$ which is quite good up to $x=5$.
Using series reversion (as @Gary commented)
$$x=1+\frac{1}{k}+\frac{1-\gamma }{k^2}+\frac{2 \left(\gamma _1+1-2 \gamma +\gamma
^2\right)}{k^3}+O\left(\frac{1}{k^4}\right)$$
Edit
If I had to solve the equation for $x$, I would prefer to solve instead
$$-\frac{\zeta (x)-1}{\zeta '(x)}=a \qquad \text{with} \qquad a=\frac 1k$$ which is much better conditioned.
A nonlinear regression gives (the coefficients were made rational), with $R^2 > 0.999999$, as an estimate,
$$x_0=\frac{1-\frac{65 }{92}a-\frac{82 }{273}a^2+\frac{27 }{125}a^3} { 1-\frac{171 }{101}a+\frac{88 }{95}a^2-\frac{16 }{99}a^3}$$
Some results
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{solution} \\
1 & 2.91932 & 2.91746 \\
2 & 1.64084 & 1.64232 \\
3 & 1.38780 & 1.38992 \\
4 & 1.27823 & 1.28026 \\
5 & 1.21697 & 1.21881 \\
6 & 1.17784 & 1.17949 \\
7 & 1.15067 & 1.15216 \\
8 & 1.13070 & 1.13205 \\
9 & 1.11541 & 1.11664 \\
10 & 1.10332 & 1.10445
\end{array}
\right)$$
Update
Another possible solution is to build the $[3,3]$ Padé approximant around $x=1$ and, making all coefficients rational have
$$f(x)=-\frac{\zeta (x)-1}{\zeta '(x)}\sim \frac {t+\frac{157 }{1308}t^2+\frac{2 }{309}t^3} {1+\frac{260 }{479}t+\frac{29 }{321}t^2+\frac{1}{224}t^3}=g(x)\qquad \text{where} \qquad t=x-1$$ which is quite good for $1 \leq x \leq 20$. To give an idea
$$\Phi=\int_0^{10} \Big[f(x)-g(x)\Big]^2\,dx=6.35 \times 10^{-6}$$
$$\Phi=\int_0^{20} \Big[f(x)-g(x)\Big]^2\,dx=4.56 \times 10^{-4}$$
Using $g(x)$, we then just need to solve the cubic equation in $t$
$$-1+\left(k-\frac{260}{479}\right) t+\left(\frac{157 k}{1308}-\frac{29}{321}\right)
t^2+\left(\frac{2 k}{309}-\frac{1}{224}\right) t^3=0$$ The discriminant is always negative and using the hyperbolic method, we have the explicit solution for $t(k)$.
Repeating the same calculations as above
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{solution} \\
1 & 2.91750 & 2.91746 \\
2 & 1.64232 & 1.64232 \\
3 & 1.38992 & 1.38992 \\
4 & 1.28026 & 1.28026 \\
5 & 1.21881 & 1.21881 \\
6 & 1.17949 & 1.17949 \\
7 & 1.15216 & 1.15216 \\
8 & 1.13205 & 1.13205 \\
9 & 1.11664 & 1.11664 \\
10 & 1.10445 & 1.10445
\end{array}
\right)$$
Best Answer
The Riemann zeta function has a simple pole at $s=1$, with residue $1$. If you take $x\gt0$ a real number, as I think you are, then: $$\begin{align}\lim_{s\to1}(\zeta(s)+x^{1-s}(1-s)^{-1})&=\lim_{s\to1}\left[\frac{x^{1-s}-1}{1-s}+\underset{\longrightarrow\gamma}{\underbrace{(\zeta(s)-(s-1)^{-1})}}\right]\\&\overset{L.H}=\gamma+\lim_{s\to1}\frac{(-\ln x)x^{1-s}}{-1}\\&=\ln(x)+\gamma\end{align}$$
As you expected, where the constant $C$ is the $\gamma$ the Euler-Mascheroni constant, or the first Stieltjes constant.
To see where these constants are coming from, I will reference for you an answer someone gave me a few months ago, here. No complex analysis is involved, really - you should only understand that "analytic continuation is unique" to understand why Conrad's zeta is the same as your zeta.