The intuition is that, following any history $h^t$, the continuation strategy should be a subgame perfect equilibrium $\sigma|h^t$, which allows you to predict the value of continuation play ($V(\sigma|h^t)$) with certainty. Therefore, when choosing the current-stage strategy, a player know her action $a$ counts toward $h^t$, which in turn, leads to $\sigma|h^{t-1}\cup\{a\}$, generating a certain continuation payoff $v_i(\sigma|h^{t-1}\cup\{a\})$. And because other things are already determined, you can simply write $v_i(a)$. This is precisely why the continuation payoff is in fact history-strategy dependent.

The normal form says the players are choosing today's strategy $a$ by trading off today's payoff $u_i(a)$ and (discounted) future valuation $v_i(a)$. You can use the normal form because everything can be summarised into a current-stage variable $a$.

Edits: Question (1): you are right, there're infinitely many strategies for each player. However, to deal with this repeated game, we conjecture that there is some SPE $\sigma$ following action $a$. One-shot deviation principle makes it clear you never deviate from $\sigma$. Think it as some black-box algorithm that automatically choose the best action for you forever, after your play for the current period. In this sense, any future strategies are summarized by the current action $a$, and $U_i(\sigma|(h_{t-1}, a)$ is function of only $a$ but not anything else, so you can relabel it as $v_i(a)$.

Question 2: Potentially there can be many subgame perfect equilibria, including the one that (0,0) is played every period.

Question 3: Yes. There can be infinitely many.

The intuition is that, following any history $h^t$, the continuation strategy should be a subgame perfect equilibrium $\sigma|h^t$, which allows you to predict the value of continuation play ($V(\sigma|h^t)$) with certainty. Therefore, when choosing the current-stage strategy, a player know her action $a$ counts toward $h^t$, which in turn, leads to $\sigma|h^{t-1}\cup\{a\}$, generating a certain continuation payoff $v_i(\sigma|h^{t-1}\cup\{a\})$. And because other things are already determined, you can simply write $v_i(a)$. This is precisely why the continuation payoff is in fact history-strategy dependent.

The normal form says the players are choosing today's strategy $a$ by trading off today's payoff $u_i(a)$ and (discounted) future valuation $v_i(a)$. You can use the normal form because everything can be summarised into a current-stage variable $a$.

Edits: Question (1): you are right, there're infinitely many strategies for each player. However, to deal with this repeated game, we conjecture that there is some SPE $\sigma$ being played following action $a$. One-shot deviation principle makes it clear you never deviate from $\sigma$. Think it as some black-box algorithm that automatically choose the best action for you forever, after your play for the current period. In this sense, any future strategies are summarized by the current action $a$, and $U_i(\sigma|(h_{t-1}, a)$ is function of only $a$ but not anything else, so you can relabel it as $v_i(a)$.

Question (2): Then the SPE of the repeated game contains $(0,0)$ as the first period strategy, e.g., Round 1: play $0$; Round 2: if $(0,0)$ was played in round 1, play $a_{(0,0)}$, otherwise...

Since it's an infinitely repeated game, you should have one SPE where $(0,0)$ is played every period. E.g., infinitely repeated prisoner's dilemma, you have (confess,confess) to be played ``no matter what'' as an SPE, but you can also have grim trigger -- (deny, deny) is a Nash equilibrium in your normal form, and for next period, if (deny, deny) is played, then continue (deny, deny); otherwise, (confess, confess) forever.

Question (3): Yes. From the logic above, you should see there can be infinitely many SPE.

The intuition is that, following any history $h^t$, the continuation strategy should be a subgame perfect equilibrium $\sigma|h^t$, which allows you to predict the value of continuation play ($V(\sigma|h^t)$) with certainty. Therefore, when choosing the current-stage strategy, a player know her action $a$ counts toward $h^t$, which in turn, leads to $\sigma|h^{t-1}\cup\{a\}$, generating a certain continuation payoff $v_i(\sigma|h^{t-1}\cup\{a\})$. And because other things are already determined, you can simply write $v_i(a)$. This is precisely why the continuation payoff is in fact history-strategy dependent.

The normal form says the players are choosing today's strategy $a$ by trading off today's payoff $u_i(a)$ and (discounted) future valuation $v_i(a)$. You can use the normal form because everything can be summarised into a current-stage variable $a$.

The intuition is that, following any history $h^t$, the continuation strategy should be a subgame perfect equilibrium $\sigma|h^t$, which allows you to predict the value of continuation play ($V(\sigma|h^t)$) with certainty. Therefore, when choosing the current-stage strategy, a player know her action $a$ counts toward $h^t$, which in turn, leads to $\sigma|h^{t-1}\cup\{a\}$, generating a certain continuation payoff $v_i(\sigma|h^{t-1}\cup\{a\})$. And because other things are already determined, you can simply write $v_i(a)$. This is precisely why the continuation payoff is in fact history-strategy dependent.

The normal form says the players are choosing today's strategy $a$ by trading off today's payoff $u_i(a)$ and (discounted) future valuation $v_i(a)$. You can use the normal form because everything can be summarised into a current-stage variable $a$.

Edits: Question (1): you are right, there're infinitely many strategies for each player. However, to deal with this repeated game, we conjecture that there is some SPE $\sigma$ following action $a$. One-shot deviation principle makes it clear you never deviate from $\sigma$. Think it as some black-box algorithm that automatically choose the best action for you forever, after your play for the current period. In this sense, any future strategies are summarized by the current action $a$, and $U_i(\sigma|(h_{t-1}, a)$ is function of only $a$ but not anything else, so you can relabel it as $v_i(a)$.

Question 2: Potentially there can be many subgame perfect equilibria, including the one that (0,0) is played every period.

Question 3: Yes. There can be infinitely many.