Yes, your formula is true (and no RH is needed),

Your symmetrized zeta-function is symmetrized such that the functional equation

$$
\zeta^* (s)=\zeta^* (1-s)
$$
holds. Thus we have that $\zeta^* (s)-\zeta^* (1-s)= 0 $ and on the critical line for $ s = 1/2+it $ this gives us that

$$
2 \Im \left( \zeta^* ( \frac 1 2 + it ) \right) = \zeta^* ( \frac 1 2 + it ) - \zeta^* ( \frac 1 2 - it ) = 0,
$$
i.e. $\zeta^*(1/2+it)$ is real valued (in fact it equals the Hardy $Z$-function).

Taking the derivative of both sides gives us $ { \zeta^* }' (s)= - { \zeta^* }' (1-s) $ and thus
$ { \zeta^* }' (s) + { \zeta^* } '(1-s)=0 $. On the critical line this gives us

$$
2 \Re \left( { \zeta^* } '( \frac 1 2+it) \right) = { \zeta^* } '( \frac 1 2 + it ) + { \zeta^* } '( \frac 1 2 -it )=0,
$$
and thus ${\zeta^*}' (1/2+it)$ is imaginary. That answers your question (1).

Now taking the logarithmic derivative of the identity

$$
\zeta^*(s)=\pi^{-s/2} \Gamma(s/2) \zeta(s)
$$
gives us that

$$
\frac{ { \zeta^ * } '(s) } { \zeta ^ * (s) } = -(\log \pi)/2 + \frac {\Gamma'(s/2)}{2\Gamma(s/2)} + \frac{ \zeta '(s) } { \zeta (s) } $$

By the fact that the left hand side is imaginary (imaginary divided by real) on the critical line, taking the real part of the last identity yields your numerically verified result.

Since the di-gamma function $ \psi(s)=\Gamma'(s)/\Gamma(s) $ is a nice function with asymptotic expansion and by the fact that the first order term $s^{-1}$ vanishes when taking the real part we see that $ \Re( \psi(1/4+it/2)) =\log (t/2) +O(t^{-2}) $, but if we want more terms we can easily get an error term of order $t^{-N}$ for any $N>0$. Thus your formula gives for example:

$$ \Re \left( \frac{ \zeta' (\frac 1 2 +it ) } { \zeta (\frac 1 2 + it ) } \right) = - \frac 1 2 \log \left(\frac t {2 \pi} \right) +O(t^{-2}). $$