\begin{equation}
t_{\beta}^{\prime} = \gamma t_{\alpha} + \frac{v\gamma t_{\alpha}}{c}
\tag{01}
\end{equation}
Your equation (01) is right, but your equation (02) is wrong :
\begin{equation}
t_{\beta} \ne \frac{1}{\gamma}t_{\beta}^{\prime}
\tag{02}
\end{equation}
The right (02) equation is :
\begin{equation}
t_{\beta} =\gamma t_{\beta}^{\prime}
\tag{02$^{\prime}$}
\end{equation}
So, multiplying (01) by $\gamma$ yields
\begin{equation}
t_{\beta} = \gamma^{2}\left( 1 + \frac{v}{c}\right)t_{\alpha}=\dfrac{\left( 1 + \frac{v}{c}\right)}{\left(1 -\frac{v^2}{c^2}\right)}t_{\alpha}=\dfrac{1}{1-\frac{v}{c}}t_{\alpha}
\tag{03}
\end{equation}
When two events happen at the same point, $\Delta x = 0 $, at $\Delta t$ time apart in the unprimed system (here A) then in the primed system (here B) happen at time apart
\begin{equation}
\Delta t^{\prime}=\gamma\left(\Delta t-\frac{v \Delta x}{c^2}\right)
\tag{04}
\end{equation}
that is
\begin{equation}
\Delta t^{\prime}=\gamma\Delta t
\tag{05}
\end{equation}
So, suppose that the two systems A and B start at the same space point with $x=0, t=0$ for A and $x^{\prime}=0,t^{\prime}=0$ for B. System B is moving with speed $v$ towards the positive $x$ of A .
We have two events in A :
Event $\sigma $ : Observer A has observer B on $x_{\sigma}=0$ at $t_{\sigma}=0$.
Event $ \alpha $ : Observer A from $x_{\alpha}=0$ at $t=t_{\alpha}$ sends a light-pulse to B.
We have
\begin{equation}
\Delta x = x_{\alpha}-x_{\sigma}=0 , \qquad \Delta t = t_{\alpha}-t_{\sigma}=t_{\alpha}
\tag{06}
\end{equation}
So from (05)
\begin{equation}
\Delta t^{\prime} = t_{\alpha}^{\prime}-t_{\sigma}^{\prime}=\gamma \Delta t =\gamma \left(t_{\alpha}-t_{\sigma}\right)
\tag{07}
\end{equation}
that is
\begin{equation}
t_{\alpha}^{\prime}=\gamma t_{\alpha}
\tag{08}
\end{equation}
That's all right till now.
Now let the following two events in B :
Event $\sigma $ : Observer B has observer A on $x_{\sigma}^{\prime}=0$ at $t_{\sigma}^{\prime}=0$.
Event $ \beta $ : Observer B receives on $x_{\beta}^{\prime}=0$ at $t=t_{\beta}^{\prime}$ the light-pulse from A.
We have
\begin{equation}
\Delta x ^{\prime}= x_{\beta}^{\prime}-x_{\sigma}^{\prime}=0 , \qquad \Delta t ^{\prime}= t_{\beta}^{\prime}-t_{\sigma}^{\prime}= t_{\beta}^{\prime}
\tag{09}
\end{equation}
Of course here we have for the inverse Lorentz transformation
\begin{equation}
\Delta t=\gamma\left(\Delta t ^{\prime}+\frac{v \Delta x^{\prime}}{c^2}\right)
\tag{10}
\end{equation}
So,
\begin{equation}
\Delta t = t_{\beta}-t_{\sigma}=\gamma \Delta t ^{\prime}= \gamma \left( t_{\beta}^{\prime}-t_{\sigma}^{\prime}\right)= \gamma t_{\beta}^{\prime}
\tag{11}
\end{equation}
and finally
\begin{equation}
t_{\beta} =\gamma t_{\beta}^{\prime}
\tag{02$^{\prime}$}
\end{equation}
When two events happen at the same point of an unprimed system (here A) then in the primed system (here B)
\begin{equation}
\Delta t^{\prime}=\gamma\Delta t
\tag{05}
\end{equation}
When two events happen at the same point of an primed system (here B) then in the unprimed system (here A)
\begin{equation}
\Delta t=\gamma\Delta t^{\prime}
\tag{12}
\end{equation}