The following task is given to me in terms of numerics:
$$ \alpha_{t}=\frac{\left\|g^{t}\right\|_{2}^{2}}{\left(d^{t}, A d^{t}\right)}, \quad \beta_{t}=\frac{\left\|g^{t+1}\right\|_{2}^{2}}{\left\|g^{t}\right\|_{2}^{2}} $$
I had the following in thoughts (edit):
According to Lemma 2.18 (iii), $g^{t}$ is orthogonal to $d^{i}$ for $i \leq t-1$, and thus $\left\langle g^{t}, d^{i}\right\rangle=0$ for $i \leq t-1$. Since $d^{t} \in-g^{t}+\operatorname{span}\left(d^{0}, d^{1}, \ldots, d^{t-1}\right)$ and is therefore a linear combination of $g^{t}$ and the previous $d^{i}$: $d^{t}=-g^{t}+\sum \limits_{i=0}^{t-1} \beta_{i} d^{i}$
Now $\left\langle g^{t}, d^{t}\right\rangle=-\left\langle g^{t}, g^{t}\right\rangle$, since $g^{t}$ is orthogonal to span $ \left(d^{0}, d^{1}, \ldots, d^{t-1}\right)$: $$ \alpha_{t}=\frac{\left\|g^{t}\right\|_{2}^{2}}{\left\langle d^{t}, A d^{t}\right\rangle} $$
To show beta, I used Lemma 2.18 (iv) and that all $g^{t}$ are orthogonal to each other, i.e. $\left\langle g^{t}, g^{i}\right\rangle=0$ for $i \neq t$:
$\begin{array}{l}\beta_{t}=\frac{\left\langle g^{t+1} A d^{t}\right\rangle}{\left\langle d^{t}, A d^{t}\right\rangle}=\left\langle g^{t+1},-\frac{\left\langle g^{t}, d^{t}\right\rangle}{\left\langle d^{t}, A d^{t}\right\rangle} A d^{t}\right\rangle \frac{1}{-\left\langle g^{t}, d^{t}\right\rangle}=\left\langle g^{t+1}, g^{t}-\frac{\left\langle g^{t}, d^{t}\right\rangle}{\left\langle d^{t}, A d^{t}\right\rangle} A d^{t}\right\rangle \frac{1}{\left\|g^{t}\right\|_{2}^{2}}=\left\langle g^{t+1}, g^{t}-\alpha_{t} A d^{t}\right\rangle \frac{1}{\left\|g^{t}\right\|_{2}^{2}}= \\ \quad \frac{\left\langle g^{t+1}, g^{t+1}\right\rangle}{\left\|g^{t}\right\|_{2}^{2}}=\frac{\left\|g^{t+1}\right\|_{2}^{2}}{\left\|g^{t}\right\|_{2}^{2}}\end{array}$
I have used the following:
Conjugate Gradient Method Determine descent directions that are pairwise orthogonal to each other. Initialization: $x^{0} \in \mathbb{R}^{n}, d^{0}:=-g^{0}=b-A x^{0}$ Iteration: $$ \begin{aligned} t & =0,1,2, \ldots \\ x^{t+1} & =x^{t}+\alpha_{t} d^{t}, \alpha_{t}=-\frac{\left(g^{t}, d^{t}\right)}{\left(d^{t}, A d^{t}\right)} \\ g^{t+1} & =g^{t}+\alpha_{t} A d^{t}\quad (g^{t+1}=A x^{t+1}-b=\underbrace{A x^{t}-b}_{g^{t}}+\alpha_{t} A d^{t}) \\ d^{t+1} & =-g^{t+1}+\beta_{t} d^{t}, \beta_{t}=\frac{\left(g^{t+1}, A d^{t}\right)}{\left(d^{t}, A d^{t}\right)} \end{aligned} $$
Lemma 2.18 Suppose $g^{i} \neq 0, i \leq t \in \mathbb{N}$, then:
(i) $\operatorname{span}\left(g^{0}, \ldots, g^{t}\right)=\operatorname{span}\left(A^{0} g^{0}, \ldots, A^{t} g^{0}\right)$
(ii) $\operatorname{span}\left(g^{0}, \ldots, g^{t}\right)=\operatorname{span}\left(d^{0}, \ldots, d^{t}\right)$
(iii) $\left(g^{t}, d^{i}\right)=0, i \leq t-1$
(iv) $\left(d^{t}, A d^{i}\right)=0, i \leq t-1$
Remark 2.19
(i) According to Lemma 2.18 iv), the descent directions are pairwise "$A$-orthogonal" (or $A$-conjugate).
(ii) The descent directions are linearly independent, and it holds that $\operatorname{span}\left(d^{0}, \ldots, d^{n-1}\right)=\mathbb{R}^{n}$ if $g^{i} \neq 0, i \leq n-1$.
(iii) From Lemma 2.18 iii) and iv), it follows that $$ \alpha_{t}=\frac{\left\|g^{t}\right\|^{2}}{\left(d^{t}, A d^{t}\right)}, \beta_{t}=\frac{\left\|g^{t+1}\right\|^{2}}{\left\|g^{t}\right\|^{2}} $$
Use $d^t \in - g^t + \text{span}(d^0, d^1, \cdots, d^{t-1})$.
Then $\langle{d^t, g^t\rangle} = - \langle g^t, g^t\rangle$. Because $g^t$ is orthogonal to the space on RHS, see your (iii).
Similar idea applies to $\beta$, use your (iv).