The first thing in the diagram are the solid arrows: those are maps that you "know."
You "know" the map $f\colon R\to S$, because you are assuming that you are given this map.
You also "know" the map $\pi\colon R\to R/\mathrm{ker}(f)$. This is simply the canonical projection onto the quotient. Remember that whenever you have a ring $R$ and an ideal $\mathcal{I}$ of $R$, you can form the quotient ring $R/\mathcal{I}$; and there is a canonical projection map $\pi\colon R\to R/\mathcal{I}$ given by $\pi(r) = r+\mathcal{I}$. In this case, since $f$ is a homomorphism, you know that $\mathrm{ker}(f)$ is an ideal, so you automatically get the quotient ring $R/\mathrm{ker}(f)$ and the quotient map $\pi\colon R\to R/\mathrm{ker}(f)$.
The second thing is that the "dashed arrow" represents a homomorphism which the theorem asserts exists; so one thing the theorem asserts is that there is a ring homomorphism $R/\mathrm{ker}(f)\to S$, and which is an isomorphism (assuming $f$ is onto).
Finally, saying the diagram "is commutative" means the following: whenever you have two paths from one ring in the diagram to another (in this case, there are two ways to get from $R$ to $S$: either through the arrow $f$, or by first going to $R/\mathrm{ker}(f)$ via the map I called $\pi$, and the map $R/\mathrm{ker}(f)\to S$ that we are asserting exists), then the results you get are the same no matter which path you take (in this case: if $r\in R$, then $f(r)$ will be the same as first taking $\pi(r)$, and then applying the map $R/\mathrm{ker}(f)\to S$ to $\pi(r)$.
For an example, take $R=\mathbb{Z}\times\mathbb{Z}$, $S=(\mathbb{Z}/10\mathbb{Z})\times(\mathbb{Z}/3\mathbb{Z})$, and $f(a,b) = (a\bmod 10, b\bmod 3)$.
This is a ring homomorphism: \begin{align*} f\bigl((a,b) + (c,d)\bigr) &= f(a+c,b+d) = \bigl( a+c\bmod 10, b+d\bmod 3\bigr)\\ &= (a\bmod 10,b\bmod 3) + (c\bmod 10, d\bmod 3) = f(a,b) + f(c,d),\\ f\bigl((a,b)(c,d)\bigr) &= f(ac,bd) = \bigl(ac\bmod 10, bd\bmod 3\bigr)\\ &= (a\bmod 10, b\bmod 3)(c\bmod 10,d\bmod 3) = f(a,b)f(c,d). \end{align*} It is also onto: given $(x\bmod 10,y\bmod 3)\in S$, we can always find an element $(a,b)\in R$ such that $f(a,b)=(x\bmod 10, y\bmod 3)$. For instance, we can take $a=x$, $b=y$.
what is the kernel of this map? $f(a,b) = (0\bmod 10,0\bmod 3)$ if and only if $a\equiv 0 \pmod{10}$ and $b\equiv 0 \pmod{3}$. So the kernel is $10\mathbb{Z}\times 3\mathbb{Z}$.
Now, we have a map from $R$ to $R/\mathrm{ker}(f)$; namely, \begin{align*} \pi\colon \mathbb{Z}\times\mathbb{Z} &\longmapsto \frac{\mathbb{Z}\times\mathbb{Z}}{10\mathbb{Z}\times 3\mathbb{Z}}\\ (a,b) &\longmapsto (a,b) + 10\mathbb{Z}\times 3\mathbb{Z} \end{align*} What the theorem states is that there is an isomorphism $\varphi\colon \frac{\mathbb{Z}\times\mathbb{Z}}{10\mathbb{Z}\times3\mathbb{Z}} \to \left(\frac{\mathbb{Z}}{10\mathbb{Z}}\right)\times\left(\frac{\mathbb{Z}}{3\mathbb{Z}}\right)$ which makes "the diagram commute". That is, for every $(a,b)\in R$, we will have $f(a,b) = \phi(\pi(a,b)).$ What is $\phi$? $\phi\bigl((a,b) + 10\mathbb{Z}\times 3\mathbb{Z}\bigr) = (a\bmod 10,b\bmod 3).$ Does this work? First, $\phi$ is well defined: if $(a,b)+10\mathbb{Z}\times 3\mathbb{Z} = (c,d)+10\mathbb{Z}\times3\mathbb{Z}$, then $(a,b)-(c,d)\in 10\mathbb{Z}\times 3\mathbb{Z}$; this means that $a-c\in 10\mathbb{Z}$ and $b-d\in 3\mathbb{Z}$, so $a\bmod 10 = c\bmod 10$ and $b\bmod 3$ = $d\bmod 3$. Thererfore, $(a\bmod 10,b\bmod 3) = (c\bmod 10, d\bmod 3)$. Thus, $\phi\bigl( (a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr) = \phi\bigl( (c,d)+10\mathbb{Z}\times 3\mathbb{Z}\bigr),$ which shows that $\phi$ is well defined.
It is also straightforward to check that $\phi$ is a ring homomorphism, and that $f(a,b) = \phi(\pi(a,b))$ for all $(a,b)\in R$. To show that $\phi$ is an isomorphism, note that it must be a surjection (because $\phi\circ\pi = f$ is a surjection, and if a composition is surjective, then the second map is surjective), and if $(0\bmod 10,0\bmod 3) = \phi\bigl( (a,b) + 10\mathbb{Z}\times 3\mathbb{Z}\bigr) = (a\bmod 10, b\bmod 3),$ then $a\equiv 0\pmod{10}$ and $b\equiv 0 \pmod{3}$, so $(a,b)\in 10\mathbb{Z}\times 3\mathbb{Z}$; this means that $(a,b)+ 10\mathbb{Z}\times 3\mathbb{Z} = (0,0)+10\mathbb{Z}\times3\mathbb{Z},$ so the map is one-to-one. Thus, $\phi$ is an isomorphism, as desired.
Also, $\phi$ is the only homomorphism that "fits" into the diagram: because if $\psi$ also "fits", then given $(a,b)+10\mathbb{Z}\times 3\mathbb{Z}$, we can write this as $\pi(a,b)$, so $\psi\bigl((a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr) = \psi(\pi(a,b)) = f(a,b) = \phi(\pi(a,b)) = \phi\bigl( (a,b)+10\mathbb{Z}\times 3\mathbb{Z}\bigr),$ so $\psi=\phi$.
"Up to isomorphism." What about your final question? Consider the integers "in English" and the integers "in Spanish" ("los enteros"). The names of the elements are different ("one", "two", etc., vs. "uno", "dos", etc.) But "the integers" and "los enteros" are essentially the same ring: it's just a question of what we call the elements, not how we add them or multiply them.
Explicitly, we have a "translation" betwen "the integers" and "los enteros." This translation is such that if you take two integers, add them, and then translate the answer, you get the same thing as if you first translate the integer and then add them in Spanish. And the same thing with multiplication. Even though "the integers" and "los enteros" are technically different things (as sets), if all we are concerned about is the ring-theoretic properties of these rings, then they are essentially "the same ring." We don't really care whether we call the unit element "one" or "uno", the important point is that it is a unit, that if you add it to itself you will never get the zero element ("zero" or "cero", depending on which ring you are in) etc. The translation from English to Spanish and the translation from Spanish to English lets us go back and forth between the two at will, and none of the ring-theoretic properties will depend on which language we are using, but only on the ring-theoretic properties of the elements.
So, for all ring theoretic mathematical purposes, "the integers" are the same as "los enteros": because we have a perfect translation between the two rings, namely, an isomorphism. So, the two are really "equivalent" (for any ring-theoretic investigation you might want to undertake).
This means that there is really no point in distinguishing between the integers and los enteros. They are "equivalent rings", because they is an isomorphism between them. We say this by saying that they are "equivalent up to isomorphism."
(More formally: if you imagine the collection of all rings (okay, there are foundational problems with that, but ignore them; they are easy to get around), then the relation $R\cong S$ given by "there is a ring isomorphism between $R$ and $S$" is an equivalence relation: it is reflexive (the identity map shows $R\cong R$); symmetric (if $R\cong S$, then there is an isomorphism $f\colon R\to S$, so $f^{-1}\colon S\to R$ is an isomorphism from $S$ to $R$, proving $S\cong R$); and transitive: if $R\cong S$ and $S\cong T$, then there exist $f\colon R\to S$ and $g\colon S\to T$ that are isomorphisms, and $g\circ f\colon R\to T$ is an isomorphism, showing $R\cong T$. So we have an equivalence relation between rings, which partitions the collection of all rings into equivalence classes. When two rings are in the same equivalence class, we say they are "equivalent up to isomorphism", because the equivalence relation is "there is an isomorphism between".)