Since the region $D:\{(x,y) ~|~ 0 \leq y \leq x, 0 \leq x \leq 1\}$, then the integral is simply
$$
\begin{align}
\iint_D (x + y)\ dx\ dy&=\int_{x=0}^1\int_{y=0}^x (x+y)\ dy\ dx\\
&=\int_{x=0}^1\left[xy+\frac{1}{2}y^2\right]_{y=0}^x \ dx\\
&=\int_{x=0}^1\left(\frac{3}{2}x^2\right)\ dx\\
&=\left[\frac{1}{2}x^3\right]_{x=0}^1\\
&=\boxed{\color{blue}{\Large\frac{1}{2}}}
\end{align}
$$
No transformation of variables needed.
UPDATE:
If we want to answer this question by using transformation of variables: $x=u+v$ and $y=u-v$, then the region $D$ in $uv$-coordinate corresponds to the region $$0\le y\le x\;\Rightarrow\;0\le u-v\le u+v\;\Rightarrow\;-v\le v\le u$$ and $$0\le x\le 1\;\Rightarrow\;0\le u+v\le 1.$$ Take a look the picture below.
Thus,
$$
\begin{align}
\iint_D (x + y)\ dx\ dy&=\iint_D 2u\ |J|\ du\,dv\\
&=4\iint_D u \ dv\,du\\
&=4\left(\int_{u=0}^\frac{1}{2}\int_{v=0}^u u\ dv\,du+\int_{u=\frac{1}{2}}^1\int_{v=0}^{1-u} u\ dv\,du\right)\\
&=4\left(\int_{u=0}^\frac{1}{2} u^2\ du+\int_{u=\frac{1}{2}}^1 u(1-u)\ du\right)\\
&=4\left(\left.\frac{1}{3} u^3\right|_{u=0}^\frac{1}{2}+\left.\frac{1}{2} u^2-\frac{1}{3} u^3\right|_{u=\frac{1}{2}}^1\right)\\
&=4\left(\frac{1}{24}+\frac{3}{8} -\frac{7}{24}\right)\\
&=\boxed{\color{blue}{\Large\frac{1}{2}}}
\end{align}
$$
$$\\$$
$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
Suppose $\mu$ and $\gamma$ are probability measures on $(X,\mathscr{F})$, $\mu\ll\gamma$, and $T:(X,\mathscr{F})\rightarrow(Y,\mathscr{G})$ measurable.
Then of course $\mu\circ T^{-1}\ll\gamma\circ T^{-1}$, for $\gamma\circ T^{-1}(A)=\gamma(T^{-1}(A))=0$, implies $\mu(T^{-1}(A))=\mu\circ T^{-1}(A)=0$.
Claim:
$$ \mathbb{E}_\gamma\Big[\frac{d\mu}{d\gamma}\big|\sigma(T)\Big]=\frac{d(\mu\circ T^{-1})}{d(\gamma\circ T^{-1})}\circ T$$
Let $h:(Y,\mathscr{G})\mapsto(\mathbb{R},\mathscr{B}(\mathbb{R})$ be a measurable function such that $E_\gamma\Big[\frac{d\mu}{d\gamma}\big|\sigma(T)\Big]=h\circ T$ (any function $\phi$ that is measurable with respect to $\sigma(T)$ admits a representation for the form $\phi=h_\phi\circ T$ for some measurable function $h$ on $Y$). Then, for any $B\in\mathscr{G}$,
$$\begin{align}
\int_Y \mathbb{1}_B\,\frac{d(\mu\circ T^{-1})}{d(\gamma\circ T^{-1})}\, d(\gamma\circ T^{-1})&=\int_Y \mathbb{1}_B\,d(\mu\circ T^{-1})=\int_X \mathbb{1}_B\circ T\,d\mu\\
&=\int_X\mathbb{1}_{T^{-1}(B)}\frac{d\mu}{d\gamma}\,d\gamma=\int_X\big(\mathbb{1}_{B}\circ T \big)\,\mathbb{E}_\gamma\Big[\frac{d\mu}{d\gamma}\big|\sigma(T)\Big]\,d\gamma\\
&=\int_X \big(\mathbb{1}_B\circ T\big)\, h\circ T\,d\gamma =\int_Y\mathbb{1}_B\,h\,d(\gamma\circ T^{-1})
\end{align}
$$
This proves that (1) $(\gamma\circ T^{-1})$-almost surely $\frac{d(\mu\circ T^{-1})}{d(\gamma\circ T^{-1})}=h$, and so, (2) $\frac{d(\mu\circ T^{-1})}{d(\gamma\circ T^{-1})}\circ T=\mathbb{E}_\gamma\big[\frac{d\mu}{d\gamma}\big|\sigma(T)\big]$
$\Box$
Let $\eta(x)=x \log(x)\mathbb{1}_{(0,\infty)}(x)$ on $[0,\infty)$. It is easy to check that $\eta$ is convex on $[0,\infty)$ , and that for any pair of measures $\mu$, $\gamma$ with $\mu\ll\gamma$
$$H(\mu|\gamma):=\int_X\log\big(\frac{d\mu}{d\gamma}\big)\,d\mu=\int_X\log\big(\frac{d\mu}{d\gamma}\big)\,\frac{d\mu}{d\gamma}\,d\gamma=\int_X\eta\big(\frac{d\mu}{d\gamma}\big)\,d\gamma$$
Finally, applying Jensen's inequality to conditional expectations yields
$$\begin{align}
H(\mu\circ T^{-1}|\lambda\circ T^{-1})&=\int_Y\eta\left(\frac{d\mu\circ T^{-1}}{d\gamma\circ T^{-1}}\right)\,d(\gamma\circ T^{-1})\\
&=\int_X\eta\Big(\frac{d\mu\circ T^{-1}}{d\gamma\circ T^{-1}}\circ T\Big)\,d\gamma\\
&=\int_X\eta\Big(\mathbb{E}_\gamma\big[ \frac{d\mu}{d\gamma}\big|\sigma(T)\big]\Big)\,d\gamma\\
&\leq\int_X\mathbb{E}_\gamma\big[ \eta\big(\frac{d\mu}{d\gamma}\big)\big|\sigma(T)\big]\,d\gamma\\
&=\int_X\eta\big(\frac{d\mu}{d\gamma}\big)\,d\gamma=H(\mu|\gamma)
\end{align}$$
which is the desired inequality.
Best Answer
Here's my take on it. Let $\lambda$ denote the Lebesgue measure on $\mathbb{R}^n$. The function $$ t\mapsto \nu(t)=-\int_{g(x)\geq t}\,\mathrm dx=-\lambda(\{g\geq t\}) $$ is an increasing right-continuous function on $\mathbb{R}$, and therefore gives rise to a Lebesgue-Stieltjes measure $\mu$ characterized by $$ \mu(]a,b])=\nu(b)-\nu(a),\quad a,b\in\mathbb{R},\;a<b. $$ What I take the integral $\int f(t)\,\mathrm d\nu(t)$ to mean is exactly $\int f\,\mathrm d\mu$. So we have to prove that $$ \int f\,\mathrm d\mu=\int f\circ g\,\mathrm d\lambda $$ but since $\int f\circ g\,\mathrm d\,\lambda = \int f\,\mathrm d (\lambda\circ g^{-1})$ it is enough to show that $\mu=\lambda\circ g^{-1}$. So let $a,b\in\mathbb{R}$, $a<b$, then
$$ \mu(]a,b])=\nu(b)-\nu(a)=\lambda(\{g\geq a\})-\lambda(\{g\geq b\})=\lambda(\{g\in [a,b[\})=\lambda\circ g^{-1}([a,b[), $$ but here I am stuck unfortunately. I really want to conclude that $\lambda\circ g^{-1}([a,b[)=\lambda\circ g^{-1}(]a,b])$, but I don't think that this is true for a general $g$. Maybe the assumptions you are mentioning will ensure this?