There is no joint density function since the random variable $(U,V,W,Z)$ takes values on a subset $D=\{(f_1^{-1}(x),f_2^{-1}(x),f_3^{-1}(x),f_4^{-1}(x))\mid x\in\mathbb R\}$ of $\mathbb R^4$ which has Lebesgue measure zero.
Informally, $D$ has co-dimension $3$, hence one can compare $D$ to a line in $\mathbb R^4$.
Formally, for every measurable function $\varphi$ on $\mathbb R^4$,
$$
\mathrm E(\varphi(U,V,W,Z))=\int\varphi(f_1(x),f_2(x),f_3(x),f_4(x))\,g(x)\mathrm dx,
$$
where $g$ is the density of the distribution of $X$ hence $\mathrm E(\varphi(U,V,W,Z))$ is an integral on (a subset of) $\mathbb R$ instead of $\mathbb R^4$.
The simplest analogue is when $U=V=X$ with $X$ uniformly distributed on $[0,1]$. Then $(U,V)$ is uniformly distributed on the diagonal $\Delta=\{(x,x)\mid x\in[0,1]\}$ hence the distribution of $(U,V)$ is
$$
\mathrm dP_{(U,V)}(u,v)=\mathbf 1_{u\in[0,1]}\,\delta_u(\mathrm dv)\,\mathrm du,
$$
where, for every $u$, $\delta_u$ is the Dirac distribution at $u$. One sees that $\mathrm dP_{(U,V)}(u,v)$ has no density with respect to Lebesgue measure $\mathrm du\mathrm dv$.
Your first two expressions are correct. Then
\begin{align}
\textsf{Pr}(X=x\mid Y=y)
&=\frac{\textsf{Pr}(X=x\land Y=y)}{\textsf{Pr}(Y=y)}\\
&=\frac{\frac{n!}{x!y!(n-x-y)!} p^xq^y(1-p-q)^{n-x-y}}{\frac{n!}{y!(n-y)!}q^y(1-q)^{n-y}}\\
&=\frac{(n-y)!}{x!(n-x-y)!}\frac{p^x(1-p-q)^{n-x-y}}{(1-q)^{n-y}}\\
&=\binom{n-y}x\left(\frac p{1-q}\right)^x\left(\frac{1-p-q}{1-q}\right)^{n-y-x}\;,
\end{align}
which is the binomial distribution for $n-y$ events known not to be events of type $Y$.
Best Answer
We have :
$$ P(X = m, Y = n) = \underbrace{P(X = m)}_{\text{marginal}}\underbrace{P(Y = n | X = m)}_{\text{conditional}} $$
The marginal is given, the conditional is given, thus the joint is recovered. This step is often skipped even in elementary textbooks, so it is good you asked.