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$\newcommand\R{\Bbb R}$Counterexample to Property 3: $f(x,y)=x^2(x-1)^2 y^2(y-1)^2$ for $(x,y)\in U:=\Bbb R^2$. Then $f$ is a smooth function but a connected component of $\partial\Xi_f$ contains the boundary (say $B$) of the open unit square (say $S$), which is not a smooth curve.

Indeed, take any convex function $h$ such that $h\le f$. Since $f=0$ on $B$, we have $h\le0$ on $B$ and hence on $S$. So, $\operatorname{conv}f\le0$ on $S$. On the other hand, $f>0$ on $S$, so that $S\subseteq\Xi_f$. Since the constant zero function is convex and $\le f$ and $=f$ on $B$, we see that $\operatorname{conv}f=0=f$ on $B$. So, $B\cap\Xi_f=\emptyset$. So, $B=\partial S\subseteq\partial\Xi_f$.

Property 1 is obvious. Indeed, letting $q_{a,b}(x):=q(x)+a\cdot x+b$ for any function $q$, we see that $h_{a,b}$ is a convex function such that $h_{a,b}\le f_{a,b}$ iff $h$ is a convex function such that $h\le f$, so that $\operatorname{conv}(f_{a,b})=(\operatorname{conv}f)_{a,b}$, and hence $f_{a,b}-\operatorname{conv}f_{a,b}=f-\operatorname{conv}f$.

$\newcommand\R{\Bbb R}$Counterexample to Property 3: $f(x,y)=x^2(x-1)^2 y^2(y-1)^2$ for $(x,y)\in U:=\Bbb R^2$. Then $f$ is a smooth function, but a connected component of $\partial\Xi_f$ contains the boundary (say $B$) of the open unit square (say $S$), which is not a smooth curve.

Indeed, take any convex function $h$ such that $h\le f$. Since $f=0$ on $B$, we have $h\le0$ on $B$ and hence on $S$. So, $\operatorname{conv}f\le0$ on $S$. On the other hand, $f>0$ on $S$, so that $S\subseteq\Xi_f$. Since the constant zero function is convex and $\le f$ and $=f$ on $B$, we see that $\operatorname{conv}f=0=f$ on $B$. So, $B\cap\Xi_f=\emptyset$. So, $B=\partial S\subseteq\partial\Xi_f$.

For an illustration, here is the graph $\{(x,y,f(x,y))\colon(x,y)\in[-0.2,1.2]^2\}$:

Property 1 is obvious. Indeed, letting $q_{a,b}(x):=q(x)+a\cdot x+b$ for any function $q$, we see that $h_{a,b}$ is a convex function such that $h_{a,b}\le f_{a,b}$ iff $h$ is a convex function such that $h\le f$, so that $\operatorname{conv}(f_{a,b})=(\operatorname{conv}f)_{a,b}$, and hence $f_{a,b}-\operatorname{conv}f_{a,b}=f-\operatorname{conv}f$.

Counterexample to Property 3: $f(x,y)=x^2(x-1)^2 y^2(y-1)^2$ for $(x,y)\in U:=\Bbb R^2$. Then $f$ is a smooth function but a connected component of $\partial\Xi_f$ contains the boundary of the unit square, which is not a smooth curve.

Property 1 is obvious. Indeed, letting $q_{a,b}(x):=q(x)+a\cdot x+b$ for any function $q$, we see that $h_{a,b}$ is a convex function such that $h_{a,b}\le f_{a,b}$ iff $h$ is a convex function such that $h\le f$, so that $\operatorname{conv}(f_{a,b})=(\operatorname{conv}f)_{a,b}$, and hence $f_{a,b}-\operatorname{conv}f_{a,b}=f-\operatorname{conv}f$.

$\newcommand\R{\Bbb R}$Counterexample to Property 3: $f(x,y)=x^2(x-1)^2 y^2(y-1)^2$ for $(x,y)\in U:=\Bbb R^2$. Then $f$ is a smooth function but a connected component of $\partial\Xi_f$ contains the boundary (say $B$) of the open unit square (say $S$), which is not a smooth curve.

Indeed, take any convex function $h$ such that $h\le f$. Since $f=0$ on $B$, we have $h\le0$ on $B$ and hence on $S$. So, $\operatorname{conv}f\le0$ on $S$. On the other hand, $f>0$ on $S$, so that $S\subseteq\Xi_f$. Since the constant zero function is convex and $\le f$ and $=f$ on $B$, we see that $\operatorname{conv}f=0=f$ on $B$. So, $B\cap\Xi_f=\emptyset$. So, $B=\partial S\subseteq\partial\Xi_f$.

Property 1 is obvious. Indeed, letting $q_{a,b}(x):=q(x)+a\cdot x+b$ for any function $q$, we see that $h_{a,b}$ is a convex function such that $h_{a,b}\le f_{a,b}$ iff $h$ is a convex function such that $h\le f$, so that $\operatorname{conv}(f_{a,b})=(\operatorname{conv}f)_{a,b}$, and hence $f_{a,b}-\operatorname{conv}f_{a,b}=f-\operatorname{conv}f$.

Counterexample to Property 3: $f(x,y)=x^2(x-1)^2 y^2(y-1)^2$ for $(x,y)\in U:=\Bbb R^2$. Then $f$ is a smooth function but a connected component of $\partial\Xi_f$ contains the boundary of the unit square, which is not a smooth curve.

Counterexample to Property 3: $f(x,y)=x^2(x-1)^2 y^2(y-1)^2$ for $(x,y)\in U:=\Bbb R^2$. Then $f$ is a smooth function but a connected component of $\partial\Xi_f$ contains the boundary of the unit square, which is not a smooth curve.

Property 1 is obvious. Indeed, letting $q_{a,b}(x):=q(x)+a\cdot x+b$ for any function $q$, we see that $h_{a,b}$ is a convex function such that $h_{a,b}\le f_{a,b}$ iff $h$ is a convex function such that $h\le f$, so that $\operatorname{conv}(f_{a,b})=(\operatorname{conv}f)_{a,b}$, and hence $f_{a,b}-\operatorname{conv}f_{a,b}=f-\operatorname{conv}f$.