Constructing the metric:
As explained in my comments above, we want to pick semi-norms $L^p(\Omega_n)$ for "larger and larger" $\Omega_n$ in order to make sure that "locally our metric looks like $L^p$". A standard way of doing this (e.g. for Fréchet spaces) is to patch these semi-norms together in the following way
$$ d(f,g) = \sum_{n\geq 1} 2^{-n}\frac{\Vert f-g \Vert_{L^p(\Omega_n)}}{1+\Vert f-g\Vert_{L^p(\Omega_n)}}. $$
First we note that the series above does indeed converge. For $\Vert f-g\Vert_{L^p(\Omega_n)}$ finite for all $n$ we have
$$ d(f,g)\leq \sum_{n\geq 1} 2^{-n} =1 $$
as
$$ 0\leq\frac{\Vert f-g \Vert_{L^p(\Omega_n)}}{1+\Vert f-g\Vert_{L^p(\Omega_n)}}\leq 1. $$
One can convince oneself that $d$ is symmetric and satisfies the triangle inequality (this might help for the later If $d(x,y)$ is a metric, then $\frac{d(x,y)}{1 + d(x,y)}$ is also a metric). In order for this to be a metric we need that $d(f,g)=0$ implies $f=g$. In order for this to be true, we need that
$$ \bigcup_{n\geq 1} \Omega_n = \Omega. $$
Our goal is to metrize the space $L^p_\mathrm{loc}(\Omega)$, thus, we only know $\Vert f-g\Vert_{L^p(A)}<\infty$ for $A$ contained in a compact set. Hence, we need to make sure that $\Omega_n$ are contained in a compact set.
Now, if $\Omega=\mathbb{R}^n$, then we could simply take $\Omega_n=\overline{B(0,n)}$ (closed ball of radius $n$ centered at the origin). For a more general domain $\Omega\subseteq \mathbb{R}^n$ one might be tempted to just pick $\Omega_n = \Omega \cap \overline{B(0,n)}$. What is the problem? Let's take $\Omega=B(0,1)$, then we see that $B(0,1) \cap \overline{B(0,n)} = B(0,1)$ is not contained in any compact subset of $B(0,1)$. The issue is that we are getting "too close to the boundary of $\Omega$". One way of staying away from the boundary is to just consider the set
$$ A_n = \{ x\in \Omega \ : \ \mathrm{dist}(x,\partial \Omega) \geq \frac{1}{n} \}. $$
Clearly, $A_n$ is closed, never touches the boundary and $\bigcup_{n\geq 1} A_n = \Omega$ (assuming that $\Omega$ is open, alternatively we could assume that $\partial \Omega$ is a null set, then $\bigcup_{n\geq 1} A_n$ equals $\Omega$ up to potentially a null set, which is good enough for our construction). However, $A_n$ needs not be bounded (indeed, for $\Omega=\mathbb{R}^n$ we have $A_n=\mathbb{R}^n$).
The trick is to combine those two approaches. Namely, we can define
$$ \Omega_n = \{ x\in \Omega \cap \overline{B(0,n)} \ : \ \mathrm{dist}(x,\partial (B(0,n)\cap \Omega) \geq \frac{1}{n} \}. $$
This definition makes $\Omega_n$ closed and bounded (hence compact) and we still have $\bigcup_{n\geq 1} \Omega_n = \Omega$. Indeed, let $x\in \Omega$. If $\Omega= \mathbb{R}^n$, then we have already seen that it works. Otherwise, we have for every point $x\in \Omega$ that $\mathrm{dist}(x,\partial \Omega)<\infty$. Using compactness we find $y\in \partial \Omega$ such that $\mathrm{dist}(x,\partial \Omega)=\vert x -y \vert$ (Showing that the minimum distance between a closed and compact set is attained).
We have $\partial (B(0,n)\cap \Omega)\subseteq \partial B(0,n) \cup \partial \Omega$ (see Boundaries of finite intersections and unions of sets), thus, for $n\geq \vert x \vert + \mathrm{dist}(x,\partial \Omega)+1$ we have
$$ \mathrm{dist}(x,\partial(B(0,n)\cap \Omega)) = \vert x-y \vert = \mathrm{dist}(x,\partial \Omega). $$
Hence, for $n\geq \max\{\vert x \vert + \mathrm{dist}(x,\partial \Omega)+1, 1/\mathrm{dist}(x,\partial \Omega)\} $ we have $x\in \Omega_n$.
Metric induces the correct topology:
We want to show that for $(f_n)_{n\geq 1} \subseteq L^p_\mathrm{loc}(\mathbb{R}^n)$ and $f\in L^p_\mathrm{loc}(\mathbb{R}^n)$ it is equivalent to say either
$$ \lim_{n\rightarrow \infty} d(f_n,f)=0$$
or
$$ \forall K \subseteq \Omega \ \mathrm{ compact } \lim_{n\rightarrow \infty}\int_K \vert f_n -f \vert^p =0.$$
We start by showing that the later implies the former. For this we pick $\varepsilon>0$ and $M>0$ such that
$$ \sum_{k=M+1}^\infty 2^{-k} = 2^{-M}<\varepsilon/2. $$
Then we note that
\begin{align*}
d(f_n,f)&= \sum_{k\geq 1} 2^{-k}\frac{\Vert f_n-f \Vert_{L^p(\Omega_k)}}{1+\Vert f_n-f\Vert_{L^p(\Omega_k)}} \\
&\leq \sum_{k=1}^M 2^{-k}\frac{\Vert f_n-f \Vert_{L^p(\Omega_k)}}{1+\Vert f_n-f\Vert_{L^p(\Omega_k)}} + \sum_{k=M+1}^\infty 2^{-k} \\
&< \sum_{k=1}^M 2^{-k}\frac{\Vert f_n-f \Vert_{L^p(\Omega_k)}}{1+\Vert f_n-f\Vert_{L^p(\Omega_k)}} + \varepsilon/2.
\end{align*}
However, $\Omega_k \subseteq \Omega$ is compact and hence
$$ \lim_{n\rightarrow \infty} \Vert f_n - f \Vert_{L^p(\Omega_k)} =0, $$
which allows us to find $N\geq 1$ such that for all $n\geq N$ and all $k\in \{1, \dots, M\}$
$$ \Vert f_n -f\Vert_{L^p(\Omega_k)} \leq \frac{\varepsilon}{2M}. $$
Thus, for $n\geq N$ we have $d(f_n,f)<\varepsilon$ and as we can do this for every $\varepsilon>0$ we get $\lim_{n\rightarrow \infty} d(f_n,f)=0$.
For the other direction we note that every compact $K\subseteq \Omega$ is contained in some $\Omega_k$. To see this, we note that $\mathrm{dist}(K,\partial \Omega)>0$ (using again similar arguments to Showing that the minimum distance between a closed and compact set is attained). Hence, we can first pick $\ell >0$ such that $K \subseteq \overline{B(0,\ell)}$ (works as compact sets are bounded) and then we have by similar reasoning as above that for $k \geq \max\{ \ell+1, 1/\mathrm{dist}(K, \partial \Omega) \}$ that $K\subseteq \Omega_k$. We then conclude by noting that
$$\Vert f_n-f\Vert_{L^p(K)}\leq \Vert f_n-f\Vert_{L^p(\Omega_k)} =2^k 2^{-k}\Vert f_n-f\Vert_{L^p(\Omega_k)}
\leq 2^k d(f_n,f). $$
Note that this actually shows that convergence $f_n \rightarrow f$ in $L^p_\mathrm{loc}(\mathbb{R}^n)$ is equivalent to the convergence
$$ \lim_{n\rightarrow \infty} \Vert f_n - f\Vert_{L^p(\Omega_k)} $$
for every $k$.
Completeness of the metric space:
This last observation in the previous paragraph can be used to show that $(L^p_\mathrm{loc}(\mathbb{R}^n,d)$ is complete. Let $(f_n)_{n\geq 1}$ be a Cauchy sequence in $(L^p_\mathrm{loc}(\mathbb{R}^n,d)$. Then this implies that $(f_n 1_{\Omega_k})_{n\geq 1}$ is a Cauchy sequence in $(L^p(\Omega_k), \Vert \cdot \Vert_{L^p(\Omega_k)}$. As $(L^p(\Omega_k), \Vert \cdot \Vert_{L^p(\Omega_k)}$ is complete, we know that there exists $F_k \in L^p(\Omega_k)$ such that $\lim_{n\rightarrow \infty}\Vert f_n F^k\Vert_{L^p(\Omega_k)}=0.$
We note that $\Omega_k \subseteq \Omega_{k+1}$ for all $k\geq 1$. We introduce $\Omega_0:=\emptyset$. Then, we can write $\Omega$ as the following disjoint union
$$ \Omega= \bigcup_{n\geq 1} \Omega_{n} \setminus \Omega_{n-1}. $$
We define $f(x)=F^k(x)$ if $x\in \Omega_{k}\setminus \Omega_{k-1}$. It is easy to check that $\lim_{n \rightarrow \infty} \Vert f_n -f \Vert_{L^p(\Omega_k)}=0$ for all $k$ and that $f\in L^p_\mathrm{loc}(\mathbb{R}^n)$.
