$\newcommand{\mean}[1] {\left< #1 \right>}$
$\DeclareMathOperator{\D}{d\!}$
$\DeclareMathOperator{\pr}{p}$
Proof that $\beta = \frac{1}{k T}$ and that $S = k \ln \Omega$
This proof follows from only classical thermodynamics and the microcanonical ensemble.
It makes no assumptions about the analytic form of statistical entropy, and does not involve the ideal gas law.
First recall that the pressure of an individual microstate is given from mechanics as:
\begin{align}
P_i &= -\frac{\D E_i}{\D V}
\end{align}
When assuming only $P$-$V$ mechanical work, the energy of a microstate $E_i(N,V)$ is only dependent on two variables, $N$ and $V$.
For example, consider a quantum mechanical system like particles confined in a box.
Therefore, at constant composition $N$,
\begin{align}
P_i &= -\left( \frac{\partial E_i}{\partial V} \right)_N
\end{align}
In a system described by the microcanonical ensemble, there are $\Omega$ possible microstates of the system.
The energy of an individual microstate $E_i$ is likewise trivially independent of the number microstates $\Omega$ in the ensemble.
Therefore, the pressure of an individual microstate can also be expressed as
\begin{align}
P_i &= -\left( \frac{\partial E_i}{\partial V} \right)_{\Omega,N}
\end{align}
According to statistical mechanics, the macroscopic pressure of a system is given by the statistical average of the pressures of the individual microstates:
\begin{align}
P = \mean{P} &= \sum_i^\Omega \pr_i P_i
\end{align}
where $\pr_i$ is the equilibrium probability of microstate $i$.
For a microcanonical ensemble, all microstates have the same energy $E_i = E$, where $E$ is the energy of the system.
Therefore, from the fundamental assumption of statistical mechanics, all microcanonical microstates have the same probability at equilibrium
\begin{align}
\pr_i = \frac{1}{\Omega}
\end{align}
It follows that the pressure of a microcanonical system is given by
\begin{align}
P = \mean{P} &= -\sum_i^\Omega \frac{1}{\Omega} \left( \frac{\partial E_i}{\partial V} \right)_{\Omega,N} \\
&= -\frac{1}{\Omega} \sum_i^\Omega \left( \frac{\partial E}{\partial V} \right)_{\Omega,N} \\
&= -\frac{\Omega}{\Omega} \left( \frac{\partial E}{\partial V} \right)_{\Omega,N} \\
P &= -\left( \frac{\partial E}{\partial V} \right)_{\Omega,N}
\end{align}
This expression for the pressure of a microcanonical system can be compared to the classical expression
\begin{align}
P &= -\left( \frac{\partial E}{\partial V} \right)_{S,N}
\end{align}
which immediately suggests a functional relationship between entropy $S$ and $\Omega$.
Now we take the total differential of the energy of a microcanonical system:
\begin{align}
\D E = \left(\frac{\partial E}{\partial \ln \Omega}\right)_{V, N} \D \ln\Omega + \left(\frac{\partial E}{\partial V}\right)_{\ln \Omega, N} \D V
\end{align}
As stated in the OP, for the microcanonical ensemble, the condition for thermal equilibrium is:
\begin{align}
\beta &= \left( \frac{\partial \ln \Omega}{\partial E} \right)_{V,N}
\end{align}
Thus,
\begin{align}
\D E = \frac{1}{\beta} \D \ln\Omega - P \D V
\end{align}
Compare with the classical first law of thermodynamics:
\begin{align}
\D E = T \D S - P \D V
\end{align}
Because these equations are equal, we see that
\begin{align}
T \D S &= \frac{1}{\beta} \D \ln\Omega \\
\D S &= \frac{1}{T \beta} \D \ln\Omega
\end{align}
Note that both $\D S$ and $\D \ln\Omega$ are exact differentials, so $\frac{1}{T \beta}$ must be either a constant or a function of $\Omega$.
Since $\D S$ and $\D \ln\Omega$ are both extensive quantities, $\frac{1}{T \beta}$ cannot depend on $\Omega$, and therefore
\begin{align}
k &= \frac{1}{T \beta} \\
\beta &= \frac{1}{k T}
\end{align}
where $k$ is a universal constant that is independent of composition, since $\beta$ and $T$ are both independent of composition.
By integrating, we have
\begin{align}
S &= k \ln\Omega + C
\end{align}
where $C$ is a constant that is independent of $E$ and $V$, but may depend on $N$.
By invoking the third law we can set $C=0$ to arrive at the famous Boltzmann expression for the entropy of a microcanonical system:
\begin{align}
S &= k \ln\Omega
\end{align}
The definition of temperature is
$$
T = \left(\frac{\partial S}{\partial E}\right)_{V,N}
$$
and as such it is a function of $E$, $V$ and $N$:
$$
T = T(E,V,N)
$$
The value of $T$ at a particular state is the numerical value of this function at the energy, volume and number of moles in that state.
If you have two systems you can can calculate their temperatures. If they are equal, the systems are in thermal equilibrium.
In other words, the equilibrium condition requires the value of tyemperature to be the same in both systems, not the functions:
$$
\underbrace{T(E_1,V_1,N_1)}_{T_1} =
\underbrace{T(E_2,V_2,N_2)}_{T_2}
$$
which we write more conventionally as
$$T_1 = T_2$$
I suspect the confusion arises from the common mathematical notation, $v = v(t)$, which assigns the same symbol $v$ to both the function (e.g., velocity as a function of time) and the value of the function at some specific $t$ (as in $v=0$ at $t=0$).
Best Answer
It's not unreasonable to be confused about this. Let's say you have a function $f$ of one variable and $f'$ is its derivative. Then we have $$\frac{d}{dx} f(c-x) = \color{red}{-}f'(c-x)$$ via the chain rule. If we have a function $g$ of two variables, then we might similarly write $$\frac{\partial}{\partial x} g(c-x,y) = \color{red}{-} \big(\partial_1 g\big)(c-x,y)$$ where $\partial_1g$ is the function obtained by differentiating $g$ with respect to its first entry. It is extremely common to simply call $\partial_1 g$ the same thing as $\frac{\partial g}{\partial x}$, but that only works when we assume that $x$ is the thing which we plug into the first slot of $g$. When that isn't the case - e.g. here - that notation is bad in my opinion.
In our case, we have the following expression: $$S(E,E_1) := S_1(E_1) + S_2(E - E_1)$$ $S_1(\epsilon)$ is the entropy of the first system when it has energy $\epsilon$. $S_2(\epsilon)$ is the entropy of the second system when it has energy $\epsilon$. $S(E,E_1)$ is the entropy of both systems together when they have total energy $E$, and when the first system has energy $E_1$ (so the second system has energy $E_2= E-E_1$).
If we wish to maximize this with respect to $E_1$, we would differentiate and set the result to zero: $$\frac{d}{dE_1} \big(S_1(E_1) + S_2(E-E_1)\big) = S_1'(E_1) \color{red}{-} S_2'(E-E_1) = 0$$ $$\implies S_1'(E_1) = S_2'(\underbrace{E-E_1}_{=E_2})$$
This is what your course guide is trying to say.