I am new to forcing so please forgive my mistakes. I know that one of the nice properties of countably closed forcing is that any function in $V^{\mathbb{P}}$ with domain $\omega$ is also a function in $V$. I'm not sure why this is true.
I think that the general idea is for a function $f : \check{\omega} \to \check{X}$ in $V^{\mathbb{P}}$ we find a sequence $p_0 \geq p_1 \geq \ldots$ such that $p_n \Vdash f(n) = x$. Then as $\mathbb{P}$ is countably closed we get some $q \leq p_0$ and so $q$ forces that $f$ is a function from $\check{\omega}$ to $\check{X}$. I am not sure how to find the relevant sequence of conditions. Then, once we such a $q$, how does this guarantee the function was in $V$?
I know that this question is similar to $\sigma$-closed forcing notions do not add reals but I do not know much about forcing and haven't seen how generics work.
Yes, you have the right idea.
The point is that if $f\colon\omega\to X$ (with $X\in V$) is a function in $V[G]$, and $\dot f$ is a name for $f$, then for every $n<\omega$ there is a dense set of conditions which decide the value of $f(n)$. In other words, there is a dense set of conditions, $p$, such that $p\Vdash\dot f(\check n)=\check x$. In other words, given any condition $p$, there is some $q\leq p$ and $\check x$, such that $q\Vdash\dot f(\check n)=\check x$.
So how do you "find" these $p_i$s? You don't. The axiom of choice is necessary for this proof (the statement is equivalent to the Principle of Dependent Choice), so you simply use the fact that there is an extension of $p_i$ which does the job, and you let the axiom of choice do the actual work for you.
Okay, but now we have a sequence of $p_i$, and if $q$ is a lower bound for the sequence, then we have that for every $n$ there is some $x$ such that $q\Vdash\dot f(\check n)=\check x$. So defining $g(n)=x$ if and only if $q\Vdash\dot f(\check n)=\check x$ defines a function, in $V$, and quite easily, $q\Vdash\dot f=\check g$.