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$\newcommand{\Om}{\Omega}$The answer is no.

Indeed, by rescaling, without loss of generality $L=1$. Suppose that \begin{equation} f(x_1,x_2)=g_a(|x_1-x_2|) \end{equation} for $(x_1,x_2)\in\Om$, where $g\in W^{1,2}((-2,2)^2)$ with support in $[-1,1]^2$ such that $g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)^2$.

So, if your inequality were true, then we would have \begin{equation*} L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ \le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a) \tag{1}\label{1} \end{equation*} for some real $C_1>0$ and all real $a>0$.

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.

$\newcommand{\Om}{\Omega}$The answer is no.

Indeed, by rescaling, without loss of generality $L=1$. Suppose that \begin{equation} f(x_1,x_2)=g_a(x_1-x_2) \end{equation} for $(x_1,x_2)\in\Om$, where $g_a\in W^{1,2}((-2,2)^2)$ with support in $[-1,1]^2$ such that $g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)^2$.

So, if your inequality were true, then we would have \begin{equation*} L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ \le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a) \tag{1}\label{1} \end{equation*} for some real $C_1>0$ and all real $a>0$.

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.

$\newcommand{\Om}{\Omega}$The answer is no.

Indeed, by rescaling, without loss of generality $L=1$. Suppose that \begin{equation} f(x_1,x_2)=g_a(|x_1-x_2|) \end{equation} for $(x_1,x_2)\in\Om$, where $g\in W^{1,2}((-2,2))$ with support in $[-1,1]$ such that $g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)$.

So, if your inequality were true, then we would have \begin{equation*} L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ \le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a) \tag{1}\label{1} \end{equation*} for some real $C_1>0$ and all real $a>0$.

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.

$\newcommand{\Om}{\Omega}$The answer is no.

Indeed, by rescaling, without loss of generality $L=1$. Suppose that \begin{equation} f(x_1,x_2)=g_a(|x_1-x_2|) \end{equation} for $(x_1,x_2)\in\Om$, where $g\in W^{1,2}((-2,2)^2)$ with support in $[-1,1]^2$ such that $g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)^2$.

So, if your inequality were true, then we would have \begin{equation*} L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ \le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a) \tag{1}\label{1} \end{equation*} for some real $C_1>0$ and all real $a>0$.

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.

$\newcommand{\Om}{\Omega}$The answer is no.

Indeed, by rescaling, without loss of generality $L=1$. Suppose that \begin{equation} f(x_1,x_2)=g_a(|x_1-x_2|) \end{equation} for $(x_1,x_2)\in\Om$, where $g\in W^{1,2}((-1,1))$ such that $g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)$.

So, if your were true, then we would have \begin{equation*} L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ \le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a) \tag{1}\label{1} \end{equation*} for some real $C_1>0$ and all real $a>0$.

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.

$\newcommand{\Om}{\Omega}$The answer is no.

Indeed, by rescaling, without loss of generality $L=1$. Suppose that \begin{equation} f(x_1,x_2)=g_a(|x_1-x_2|) \end{equation} for $(x_1,x_2)\in\Om$, where $g\in W^{1,2}((-2,2))$ with support in $[-1,1]$ such that $g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)$.

So, if your inequality were true, then we would have \begin{equation*} L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ \le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a) \tag{1}\label{1} \end{equation*} for some real $C_1>0$ and all real $a>0$.

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.