I'm going to start backwards and first show how the first part yields the bound on the Krull dimension of $R[t]$.
An immediate consequence is that
Any proper chain of prime ideals in $R[t]$ lying over a prime ideal in $R$ has length at most $2$.
Indeed if $Q_1, Q_2 \subsetneq P$ is such that $P_0 = Q_1 \cap R = Q_2 \cap R = P \cap R$ then we saw that $Q_1 = P_0[t] = Q_2$.
So start with a chain of maximal length in $R[t]$, $$P_1 \subsetneq P_2 \subsetneq \ldots \subsetneq P_k$$ in $R[t]$. Contraction gives us a chain $$Q_1 \subsetneq \ldots \subsetneq Q_m$$ in $R$ (after we remove redundancies), where each $Q_i = P_j \cap R$ for some $j \geq i$.
We know that each $P_i$ lies over exactly one or two $Q_j$s,
therefore $$dim(R[t]) + 1 = k \leq 2m \leq 2(dim(R) + 1)$$
Now for the justification....
Your argument that $P_0R[t] = P_0[t]$ is prime goes through fine, but I'll mention that a slick, traditional way to show this is by using the fact that $\frac{R}{M}$ is a domain iff $M$ is prime and the natural isomorphism $\frac{R[t]}{P_0[t]} \cong \left(\frac{R}{P_0}\right)[t]$.
For the main part of the question....
Start by noting that if $Q$ is $0$ then trivially $P_0 =0$ and $P_0R[t] = Q = 0$. So we can assume $0 \subsetneq Q \subsetneq P$.
Now let's break down the quotient, I'm going to change the notation to what is standard.
$RY^{-1}$ is usually referred to as the localization at $P_0$, and written $R_{P_0}$.
Now the two things we need to know about localization are: (1) for $\mathfrak{p}$ a prime ideal of $R$, we have a bijection between ideals of $R_{\mathfrak{p}}$ and ideals of $R$ contained in $\mathfrak{p}$. In particular $R_{\mathfrak{p}}$ is local ring with maximal ideal $\mathfrak{p}R_{\mathfrak{p}}$ (2) localization commutes with polynomial extension (check this by universal properties or direct computation)
Thus your quotient is
$$\frac{(R[t])_{P_0}}{(P_0 R[t])_{P_0}} \cong
\frac{R_{P_0}[t]}{P_0R_{P_0}[t]} \cong \left(\frac{R_{P_0}}{P_0R_{P_0}} \right)[t]
$$
and therefore taking $F = \frac{R_{P_0}}{P_0R_{P_0}}$ we have the claim.
Lets break this up as $$R[t] \xrightarrow{\iota} R_{P_0}[t] \xrightarrow{\pi} F[t] $$
So what are we even doing with this quotient? The core idea here is that $F[t]$ is a $PID$ so every non-zero prime ideal in $F[t]$ has height $1$. We can thus get information about $Q$ and $P$ by looking at their homomorphic images in $F[t]$ under the composition $\pi \cdot \iota$. In particular, the equality $Q=P_0R[t]$ would follow immediately if we can show that $Q \subset ker(\pi \cdot \iota)$, which would mean every polynomial in $Q$ takes coefficients in $P_0$.
Of course there are some details to work out, but the path is pretty clear. This amounts to a straightforward exercise, which i think is worth completing on your own. I'll outline details below in case that is helpful.
We start with the observation that for $\mathfrak{p}$ a prime ideal of $R[t]$ and $\mathfrak{p_0} = \mathfrak{p} \cap R$, $\mathfrak{p}R_{\mathfrak{p_0}}[t]$ is a prime ideal of $R_{\mathfrak{p_0}}[t]$ (check this for yourself). (For the special case $\mathfrak{p} \cap R = 0$, this just says $\mathfrak{p}$ extends to a prime ideal of $K[x]$ where $K$ is the quotient field of $R$.) Moreover if $\mathfrak{p} \subsetneq \mathfrak{q}$ for any ideal $\mathfrak{q} \subset R[t]$, then $\mathfrak{p}R_{\mathfrak{p_0}}[t] \subsetneq \mathfrak{q}R_{\mathfrak{p_0}}[t]$.
Thus the chain of prime ideals $$0 \subsetneq Q \subsetneq P$$ in $R[t]$ extends to a chain of prime ideals $$0 \subsetneq QR_{P_0}[t] \subsetneq PR_{P_0}[t]$$ in $R_{P_0}[t]$. Letting $\pi$ be the projection $R_{P_0}[t] \rightarrow F[t]$, we look at the chain $$0 \subset \pi(QR_{P_0}[t]) \subsetneq \pi(PR_{P_0}[t])$$ (That this is a chain of prime ideals and that the second inclusion is strict are because $\pi$ is surjective and $QR_{P_0}[t]$ contains the kernel of $\pi$). Now since this chain is in a $PID$, we deduce $\pi(QR_{P_0}[t]) = 0$, and indeed $Q \subset ker(\pi \cdot \iota)$.
