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Edit: Here's another way that gets all the $G$-theory groups. Let $X_0=\mathbb{P}^1\times \mathbb{P}^1$ and $X_1=\mathbb{P}^1\times \infty$. Then $X_0\setminus X_1\cong \mathbb{P}^1\times \mathbb{A}^1$. Now the localization sequence looks like $$\cdots \rightarrow G_{i+1}(X_0\setminus X_1)\rightarrow G_i(X_1)\rightarrow G_i(X_0)\rightarrow G_i(X_0\setminus X_1)\rightarrow G_{i-1}(X_1)\rightarrow \cdots.$$

The map $G_i(X_1)\rightarrow G_i(X_0)$ is split by the pushforward along the first projection $X_0\rightarrow \mathbb{P}^1$. This gives a decomposition $G_i(X_0)=G_i(\mathbb{P}^1)\oplus G_i(\mathbb{P}^1\times \mathbb{A}^1)$. And by homotopy invariance, which was implicit in the argument below, there is an isomorphism $G_i(\mathbb{P}^1\times \mathbb{A}^1)\cong G_i(\mathbb{P}^1)$.

We can do this using a stratification of $X=\mathbb{P}^1\times \mathbb{P}^1$. Take $X_0=X$ and $X_1=(\mathbb{P}^1\times \infty) \cup (\infty \times \mathbb{P}^1)$. Note $X_1$ is a union of two lines at a point, and $X_0\setminus X_1\cong \mathbb{A}^2$. Set $X_2=(\mathbb{P}^1\times \infty)\subset X_1$ and note $X_1\setminus X_2 \cong \mathbb{A}^1$. Lastly, set $X_3= \infty\times \infty$, which is just a point, and note $X_2\setminus X_3\cong \mathbb{A}^1$.

From the localization sequence $$G(X_3)\xrightarrow{i_3} G(X_2)\rightarrow G(X_2\setminus X_3)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_2)$ is generated by the classes $[\mathcal{O}_{X_2}]$ and $[\mathcal{O}_{X_3}]$. Similarly, from $$G(X_2)\xrightarrow{i_2} G(X_1)\rightarrow G(X_1\setminus X_2)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_1)$ is generated by $[\mathcal{O}_{X_1}], [\mathcal{O}_{X_2}]$, and $[\mathcal{O}_{X_3}]$. Lastly, $$G(X_1)\xrightarrow{i_1} G(X_0)\rightarrow G(X_0\setminus X_1)\cong \mathbb{Z}\rightarrow 0,$$ shows that $G(X_0)=G(X)$ is generated by the classes $\{[\mathcal{O}_{X_i}]\}_i$ where $i=0,1,2,3$ (in particular, $G(X)$ is a finitely generated abelian group).

Although it can be shown that $i_3$ and $i_2$ are split injections, I'm not immediately sure how to show $i_1$ is a split injection. (The morphism $i_3$ is split by the pushforward structure map $X_2\rightarrow \mathrm{Spec}(k)$, so this one is an injection, and $i_2$ is split by the pushforward along a contraction map $X_1\rightarrow X_2$.)

Instead, we can do this: let $F^i\subset G(X)$ denote the subgroup generated by classes $[\mathcal{O}_V]$ where $V$ is a subscheme of codimension $\mathrm{codim}_X(V)\geq i$. Then $F^0=G(X)$, and $F^0/F^1\cong \mathbb{Z}$ by the rank morphism; in particular, $G(X)\cong F^1\oplus \mathbb{Z}$ with $[\mathcal{O}_{X_0}]$ generating the second factor. The quotient $F^1/F^2$ is canonically the divisor class group $\mathrm{Cl}(X)$ which is isomorphic with $\mathbb{Z}\times \mathbb{Z}$ generated by a line in each fiber of the two projections $X\rightarrow \mathbb{P}^1$. This gives an isomorphism $$G(X)\cong F^1\oplus \mathbb{Z} \cong F^2\oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ Also, the map $G(X)\rightarrow F^1/F^2\cong \mathrm{Cl}(X)$ is defined so that $[\mathcal{O}_V]$ is sent to the class of the associated divisor $[V]$ for any integral subscheme $V\subset X$. Since there is an exact sequence (https://stacks.math.columbia.edu/tag/0C4J) $$0\rightarrow \mathcal{O}_{X_1}\rightarrow \mathcal{O}_{\mathbb{P}^1\times \infty}\oplus\mathcal{O}_{\infty \times \mathbb{P}^1}\rightarrow \mathcal{O}_{X_3}\rightarrow 0,$$ we see that $[\mathcal{O}_{X_1}]=[\mathcal{O}_{\mathbb{P}^1\times \infty}]+[\mathcal{O}_{\infty \times \mathbb{P}^1}] - [\mathcal{O}_{X_3}]$. It follows that $[\mathcal{O}_{X_1}]$ and $[\mathcal{O}_{X_2}]$ also generate $F^1/F^2$.

It's not too hard to see that $F^3=0$ and there is a map $F^2\rightarrow \mathbb{Z}$ gotten by pushing forward $F^2\subset G(X)\rightarrow G(\mathrm{Spec}(k))=\mathbb{Z}$, which is essentially just the degree map. The class $[\mathcal{O}_{X_3}]$ generates the image of this map and, if $K$ is the kernel $K=\mathrm{ker}(F^2\rightarrow \mathbb{Z})$ then $$G(X)=K\oplus \mathbb{Z} \oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ One can see $K=0$ a number of ways.

Edit: Here's another way that gets all the $G$-theory groups. Let $X_0=\mathbb{P}^1\times \mathbb{P}^1$ and $X_1=\mathbb{P}^1\times \infty$. Then $X_0\setminus X_1\cong \mathbb{P}^1\times \mathbb{A}^1$. Now the localization sequence looks like $$\cdots \rightarrow G_{i+1}(X_0\setminus X_1)\rightarrow G_i(X_1)\rightarrow G_i(X_0)\rightarrow G_i(X_0\setminus X_1)\rightarrow G_{i-1}(X_1)\rightarrow \cdots.$$

The map $G_i(X_1)\rightarrow G_i(X_0)$ is split by the pushforward along the first projection $X_0\rightarrow \mathbb{P}^1$. This gives a decomposition $G_i(X_0)=G_i(\mathbb{P}^1)\oplus G_i(\mathbb{P}^1\times \mathbb{A}^1)$. And by homotopy invariance, which was implicit in the argument below, there is an isomorphism $G_i(\mathbb{P}^1\times \mathbb{A}^1)\cong G_i(\mathbb{P}^1)$.

Edit: Here's an explicit description of $G_0(X)$. Everywhere below I write $G_0(-)$ as $G(-)$.

We can do this using a stratification of $X=\mathbb{P}^1\times \mathbb{P}^1$. Take $X_0=X$ and $X_1=(\mathbb{P}^1\times \infty) \cup (\infty \times \mathbb{P}^1)$. Note $X_1$ is a union of two lines at a point, and $X_0\setminus X_1\cong \mathbb{A}^2$. Set $X_2=(\mathbb{P}^1\times \infty)\subset X_1$ and note $X_1\setminus X_2 \cong \mathbb{A}^1$. Lastly, set $X_3= \infty\times \infty$, which is just a point, and note $X_2\setminus X_3\cong \mathbb{A}^1$.

From the localization sequence $$G(X_3)\xrightarrow{i_3} G(X_2)\rightarrow G(X_2\setminus X_3)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_2)$ is generated by the classes $[\mathcal{O}_{X_2}]$ and $[\mathcal{O}_{X_3}]$. Similarly, from $$G(X_2)\xrightarrow{i_2} G(X_1)\rightarrow G(X_1\setminus X_2)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_1)$ is generated by $[\mathcal{O}_{X_1}], [\mathcal{O}_{X_2}]$, and $[\mathcal{O}_{X_3}]$. Lastly, $$G(X_1)\xrightarrow{i_1} G(X_0)\rightarrow G(X_0\setminus X_1)\cong \mathbb{Z}\rightarrow 0,$$ shows that $G(X_0)=G(X)$ is generated by the classes $\{[\mathcal{O}_{X_i}]\}_i$ where $i=0,1,2,3$ (in particular, $G(X)$ is a finitely generated abelian group).

Although it can be shown that $i_3$ and $i_2$ are split injections, I'm not immediately sure how to show $i_1$ is a split injection. (The morphism $i_3$ is split by the pushforward structure map $X_2\rightarrow \mathrm{Spec}(k)$, so this one is an injection, and $i_2$ is split by the pushforward along a contraction map $X_1\rightarrow X_2$.)

Instead, we can do this: let $F^i\subset G(X)$ denote the subgroup generated by classes $[\mathcal{O}_V]$ where $V$ is a subscheme of codimension $\mathrm{codim}_X(V)\geq i$. Then $F^0=G(X)$, and $F^0/F^1\cong \mathbb{Z}$ by the rank morphism; in particular, $G(X)\cong F^1\oplus \mathbb{Z}$ with $[\mathcal{O}_{X_0}]$ generating the second factor. The quotient $F^1/F^2$ is canonically the divisor class group $\mathrm{Cl}(X)$ which is isomorphic with $\mathbb{Z}\times \mathbb{Z}$ generated by a line in each fiber of the two projections $X\rightarrow \mathbb{P}^1$. This gives an isomorphism $$G(X)\cong F^1\oplus \mathbb{Z} \cong F^2\oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ Also, the map $G(X)\rightarrow F^1/F^2\cong \mathrm{Cl}(X)$ is defined so that $[\mathcal{O}_V]$ is sent to the class of the associated divisor $[V]$ for any integral subscheme $V\subset X$. Since there is an exact sequence (https://stacks.math.columbia.edu/tag/0C4J) $$0\rightarrow \mathcal{O}_{X_1}\rightarrow \mathcal{O}_{\mathbb{P}^1\times \infty}\oplus\mathcal{O}_{\infty \times \mathbb{P}^1}\rightarrow \mathcal{O}_{X_3}\rightarrow 0,$$ we see that $[\mathcal{O}_{X_1}]=[\mathcal{O}_{\mathbb{P}^1\times \infty}]+[\mathcal{O}_{\infty \times \mathbb{P}^1}] - [\mathcal{O}_{X_3}]$. It follows that $[\mathcal{O}_{X_1}]$ and $[\mathcal{O}_{X_2}]$ also generate $F^1/F^2$.

It's not too hard to see that $F^3=0$ and there is a map $F^2\rightarrow \mathbb{Z}$ gotten by pushing forward $F^2\subset G(X)\rightarrow G(\mathrm{Spec}(k))=\mathbb{Z}$, which is essentially just the degree map. The class $[\mathcal{O}_{X_3}]$ generates the image of this map and, if $K$ is the kernel $K=\mathrm{ker}(F^2\rightarrow \mathbb{Z})$ then $$G(X)=K\oplus \mathbb{Z} \oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ One can see $K=0$ a number of ways.

We can do this using a stratification of $X=\mathbb{P}^1\times \mathbb{P}^1$. Take $X_0=X$ and $X_1=(\mathbb{P}^1\times \infty) \cup (\infty \times \mathbb{P}^1)$. Note $X_1$ is a union of two lines at a point, and $X_0\setminus X_1\cong \mathbb{A}^2$. Set $X_2=(\mathbb{P}^1\times \infty)\subset X_1$ and note $X_1\setminus X_2 \cong \mathbb{A}^1$. Lastly, set $X_3= \infty\times \infty$, which is just a point, and note $X_2\setminus X_3\cong \mathbb{A}^1$.

From the localization sequence $$G(X_3)\xrightarrow{i_3} G(X_2)\rightarrow G(X_2\setminus X_3)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_2)$ is generated by the classes $[\mathcal{O}_{X_2}]$ and $[\mathcal{O}_{X_3}]$. Similarly, from $$G(X_2)\xrightarrow{i_2} G(X_1)\rightarrow G(X_1\setminus X_2)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_1)$ is generated by $[\mathcal{O}_{X_1}], [\mathcal{O}_{X_2}]$, and $[\mathcal{O}_{X_3}]$. Lastly, $$G(X_1)\xrightarrow{i_1} G(X_0)\rightarrow G(X_0\setminus X_1)\cong \mathbb{Z}\rightarrow 0,$$ shows that $G(X_0)=G(X)$ is generated by the classes $\{[\mathcal{O}_{X_i}]\}_i$ where $i=0,1,2,3$ (in particular, $G(X)$ is a finitely generated abelian group).

Although it can be shown that $i_3$ and $i_2$ are split injections, I'm not immediately sure how to show $i_1$ is a split injection. (The morphism $i_3$ is split by the pushforward structure map $X_2\rightarrow \mathrm{Spec}(k)$, so this one is an injection, and $i_2$ is split by the pushforward along a contraction map $X_1\rightarrow X_2$.)

Instead, we can do this: let $F^i\subset G(X)$ denote the subgroup generated by classes $[\mathcal{O}_V]$ where $V$ is a subscheme of codimension $\mathrm{codim}_X(V)\geq i$. Then $F^0=G(X)$, and $F^0/F^1\cong \mathbb{Z}$ by the rank morphism; in particular, $G(X)\cong F^1\oplus \mathbb{Z}$ with $[\mathcal{O}_{X_0}]$ generating the second factor. The quotient $F^1/F^2$ is canonically the divisor class group $\mathrm{Cl}(X)$ which is isomorphic with $\mathbb{Z}\times \mathbb{Z}$ generated by a line in each fiber of the two projections $X\rightarrow \mathbb{P}^1$. This gives an isomorphism $$G(X)\cong F^1\oplus \mathbb{Z} \cong F^2\oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ Also, the map $G(X)\rightarrow F^1/F^2\cong \mathrm{Cl}(X)$ is defined so that $[\mathcal{O}_V]$ is sent to the class of the associated divisor $[V]$ for any integral subscheme $V\subset X$. Since there is an exact sequence (https://stacks.math.columbia.edu/tag/0C4J) $$0\rightarrow \mathcal{O}_{X_1}\rightarrow \mathcal{O}_{\mathbb{P}^1\times \infty}\oplus\mathcal{O}_{\infty \times \mathbb{P}^1}\rightarrow \mathcal{O}_{X_3}\rightarrow 0,$$ we see that $[\mathcal{O}_{X_1}]=[\mathcal{O}_{\mathbb{P}^1\times \infty}]+[\mathcal{O}_{\infty \times \mathbb{P}^1}] - [\mathcal{O}_{X_3}]$. It follows that $[\mathcal{O}_{X_1}]$ and $[\mathcal{O}_{X_2}]$ also generate $F^1/F^2$.

It's not too hard to see that $F^3=0$ and there is a map $F^2\rightarrow \mathbb{Z}$ gotten by pushing forward $F^2\subset G(X)\rightarrow G(\mathrm{Spec}(k))=\mathbb{Z}$, which is essentially just the degree map. The class $[\mathcal{O}_{X_3}]$ generates the image of this map and, if $K$ is the kernel $K=\mathrm{ker}(F^2\rightarrow \mathbb{Z})$ then $$G(X)=K\oplus \mathbb{Z} \oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ One can see $K=0$ a number of ways.

Edit: Here's another way that gets all the $G$-theory groups. Let $X_0=\mathbb{P}^1\times \mathbb{P}^1$ and $X_1=\mathbb{P}^1\times \infty$. Then $X_0\setminus X_1\cong \mathbb{P}^1\times \mathbb{A}^1$. Now the localization sequence looks like $$\cdots \rightarrow G_{i+1}(X_0\setminus X_1)\rightarrow G_i(X_1)\rightarrow G_i(X_0)\rightarrow G_i(X_0\setminus X_1)\rightarrow G_{i-1}(X_1)\rightarrow \cdots.$$

The map $G_i(X_1)\rightarrow G_i(X_0)$ is split by the pushforward along the first projection $X_0\rightarrow \mathbb{P}^1$. This gives a decomposition $G_i(X_0)=G_i(\mathbb{P}^1)\oplus G_i(\mathbb{P}^1\times \mathbb{A}^1)$. And by homotopy invariance, which was implicit in the above argument, there is an isomorphism $G_i(\mathbb{P}^1\times \mathbb{A}^1)\cong G_i(\mathbb{P}^1)$.

Edit: Here's another way that gets all the $G$-theory groups. Let $X_0=\mathbb{P}^1\times \mathbb{P}^1$ and $X_1=\mathbb{P}^1\times \infty$. Then $X_0\setminus X_1\cong \mathbb{P}^1\times \mathbb{A}^1$. Now the localization sequence looks like $$\cdots \rightarrow G_{i+1}(X_0\setminus X_1)\rightarrow G_i(X_1)\rightarrow G_i(X_0)\rightarrow G_i(X_0\setminus X_1)\rightarrow G_{i-1}(X_1)\rightarrow \cdots.$$

The map $G_i(X_1)\rightarrow G_i(X_0)$ is split by the pushforward along the first projection $X_0\rightarrow \mathbb{P}^1$. This gives a decomposition $G_i(X_0)=G_i(\mathbb{P}^1)\oplus G_i(\mathbb{P}^1\times \mathbb{A}^1)$. And by homotopy invariance, which was implicit in the argument below, there is an isomorphism $G_i(\mathbb{P}^1\times \mathbb{A}^1)\cong G_i(\mathbb{P}^1)$.

We can do this using a stratification of $X=\mathbb{P}^1\times \mathbb{P}^1$. Take $X_0=X$ and $X_1=(\mathbb{P}^1\times \infty) \cup (\infty \times \mathbb{P}^1)$. Note $X_1$ is a union of two lines at a point, and $X_0\setminus X_1\cong \mathbb{A}^2$. Set $X_2=(\mathbb{P}^1\times \infty)\subset X_1$ and note $X_1\setminus X_2 \cong \mathbb{A}^1$. Lastly, set $X_3= \infty\times \infty$, which is just a point, and note $X_2\setminus X_3\cong \mathbb{A}^1$.

From the localization sequence $$G(X_3)\xrightarrow{i_3} G(X_2)\rightarrow G(X_2\setminus X_3)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_2)$ is generated by the classes $[\mathcal{O}_{X_2}]$ and $[\mathcal{O}_{X_3}]$. Similarly, from $$G(X_2)\xrightarrow{i_2} G(X_1)\rightarrow G(X_1\setminus X_2)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_1)$ is generated by $[\mathcal{O}_{X_1}], [\mathcal{O}_{X_2}]$, and $[\mathcal{O}_{X_3}]$. Lastly, $$G(X_1)\xrightarrow{i_1} G(X_0)\rightarrow G(X_0\setminus X_1)\cong \mathbb{Z}\rightarrow 0,$$ shows that $G(X_0)=G(X)$ is generated by the classes $\{[\mathcal{O}_{X_i}]\}_i$ where $i=0,1,2,3$ (in particular, $G(X)$ is a finitely generated abelian group).

Although it can be shown that $i_3$ and $i_2$ are split injections, I'm not immediately sure how to show $i_1$ is a split injection. (The morphism $i_3$ is split by the pushforward structure map $X_2\rightarrow \mathrm{Spec}(k)$, so this one is an injection, and $i_2$ is split by the pushforward along a contraction map $X_1\rightarrow X_2$.)

Instead, we can do this: let $F^i\subset G(X)$ denote the subgroup generated by classes $[\mathcal{O}_V]$ where $V$ is a subscheme of codimension $\mathrm{codim}_X(V)\geq i$. Then $F^0=G(X)$, and $F^0/F^1\cong \mathbb{Z}$ by the rank morphism; in particular, $G(X)\cong F^1\oplus \mathbb{Z}$ with $[\mathcal{O}_{X_0}]$ generating the second factor. The quotient $F^1/F^2$ is canonically the divisor class group $\mathrm{Cl}(X)$ which is isomorphic with $\mathbb{Z}\times \mathbb{Z}$ generated by a line in each fiber of the two projections $X\rightarrow \mathbb{P}^1$. This gives an isomorphism $$G(X)\cong F^1\oplus \mathbb{Z} \cong F^2\oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ Also, the map $G(X)\rightarrow F^1/F^2\cong \mathrm{Cl}(X)$ is defined so that $[\mathcal{O}_V]$ is sent to the class of the associated divisor $[V]$ for any integral subscheme $V\subset X$. Since there is an exact sequence (https://stacks.math.columbia.edu/tag/0C4J) $$0\rightarrow \mathcal{O}_{X_1}\rightarrow \mathcal{O}_{\mathbb{P}^1\times \infty}\oplus\mathcal{O}_{\infty \times \mathbb{P}^1}\rightarrow \mathcal{O}_{X_3}\rightarrow 0,$$ we see that $[\mathcal{O}_{X_1}]=[\mathcal{O}_{\mathbb{P}^1\times \infty}]+[\mathcal{O}_{\infty \times \mathbb{P}^1}] - [\mathcal{O}_{X_3}]$. It follows that $[\mathcal{O}_{X_1}]$ and $[\mathcal{O}_{X_2}]$ also generate $F^1/F^2$.

It's not too hard to see that $F^3=0$ and there is a map $F^2\rightarrow \mathbb{Z}$ gotten by pushing forward $F^2\subset G(X)\rightarrow G(\mathrm{Spec}(k))=\mathbb{Z}$, which is essentially just the degree map. The class $[\mathcal{O}_{X_3}]$ generates the image of this map and, if $K$ is the kernel $K=\mathrm{ker}(F^2\rightarrow \mathbb{Z})$ then $$G(X)=K\oplus \mathbb{Z} \oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ One can see $K=0$ a number of ways.

We can do this using a stratification of $X=\mathbb{P}^1\times \mathbb{P}^1$. Take $X_0=X$ and $X_1=(\mathbb{P}^1\times \infty) \cup (\infty \times \mathbb{P}^1)$. Note $X_1$ is a union of two lines at a point, and $X_0\setminus X_1\cong \mathbb{A}^2$. Set $X_2=(\mathbb{P}^1\times \infty)\subset X_1$ and note $X_1\setminus X_2 \cong \mathbb{A}^1$. Lastly, set $X_3= \infty\times \infty$, which is just a point, and note $X_2\setminus X_3\cong \mathbb{A}^1$.

From the localization sequence $$G(X_3)\xrightarrow{i_3} G(X_2)\rightarrow G(X_2\setminus X_3)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_2)$ is generated by the classes $[\mathcal{O}_{X_2}]$ and $[\mathcal{O}_{X_3}]$. Similarly, from $$G(X_2)\xrightarrow{i_2} G(X_1)\rightarrow G(X_1\setminus X_2)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_1)$ is generated by $[\mathcal{O}_{X_1}], [\mathcal{O}_{X_2}]$, and $[\mathcal{O}_{X_3}]$. Lastly, $$G(X_1)\xrightarrow{i_1} G(X_0)\rightarrow G(X_0\setminus X_1)\cong \mathbb{Z}\rightarrow 0,$$ shows that $G(X_0)=G(X)$ is generated by the classes $\{[\mathcal{O}_{X_i}]\}_i$ where $i=0,1,2,3$ (in particular, $G(X)$ is a finitely generated abelian group).

Although it can be shown that $i_3$ and $i_2$ are split injections, I'm not immediately sure how to show $i_1$ is a split injection. (The morphism $i_3$ is split by the pushforward structure map $X_2\rightarrow \mathrm{Spec}(k)$, so this one is an injection, and $i_2$ is split by the pushforward along a contraction map $X_1\rightarrow X_2$.)

Instead, we can do this: let $F^i\subset G(X)$ denote the subgroup generated by classes $[\mathcal{O}_V]$ where $V$ is a subscheme of codimension $\mathrm{codim}_X(V)\geq i$. Then $F^0=G(X)$, and $F^0/F^1\cong \mathbb{Z}$ by the rank morphism; in particular, $G(X)\cong F^1\oplus \mathbb{Z}$ with $[\mathcal{O}_{X_0}]$ generating the second factor. The quotient $F^1/F^2$ is canonically the divisor class group $\mathrm{Cl}(X)$ which is isomorphic with $\mathbb{Z}\times \mathbb{Z}$ generated by a line in each fiber of the two projections $X\rightarrow \mathbb{P}^1$. This gives an isomorphism $$G(X)\cong F^1\oplus \mathbb{Z} \cong F^2\oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ Also, the map $G(X)\rightarrow F^1/F^2\cong \mathrm{Cl}(X)$ is defined so that $[\mathcal{O}_V]$ is sent to the class of the associated divisor $[V]$ for any integral subscheme $V\subset X$. Since there is an exact sequence (https://stacks.math.columbia.edu/tag/0C4J) $$0\rightarrow \mathcal{O}_{X_1}\rightarrow \mathcal{O}_{\mathbb{P}^1\times \infty}\oplus\mathcal{O}_{\infty \times \mathbb{P}^1}\rightarrow \mathcal{O}_{X_3}\rightarrow 0,$$ we see that $[\mathcal{O}_{X_1}]=[\mathcal{O}_{\mathbb{P}^1\times \infty}]+[\mathcal{O}_{\infty \times \mathbb{P}^1}] - [\mathcal{O}_{X_3}]$. It follows that $[\mathcal{O}_{X_1}]$ and $[\mathcal{O}_{X_2}]$ also generate $F^1/F^2$.

It's not too hard to see that $F^3=0$ and there is a map $F^2\rightarrow \mathbb{Z}$ gotten by pushing forward $F^2\subset G(X)\rightarrow G(\mathrm{Spec}(k))=\mathbb{Z}$, which is essentially just the degree map. The class $[\mathcal{O}_{X_3}]$ generates the image of this map and, if $K$ is the kernel $K=\mathrm{ker}(F^2\rightarrow \mathbb{Z})$ then $$G(X)=K\oplus \mathbb{Z} \oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ One can see $K=0$ a number of ways.

We can do this using a stratification of $X=\mathbb{P}^1\times \mathbb{P}^1$. Take $X_0=X$ and $X_1=(\mathbb{P}^1\times \infty) \cup (\infty \times \mathbb{P}^1)$. Note $X_1$ is a union of two lines at a point, and $X_0\setminus X_1\cong \mathbb{A}^2$. Set $X_2=(\mathbb{P}^1\times \infty)\subset X_1$ and note $X_1\setminus X_2 \cong \mathbb{A}^1$. Lastly, set $X_3= \infty\times \infty$, which is just a point, and note $X_2\setminus X_3\cong \mathbb{A}^1$.

From the localization sequence $$G(X_3)\xrightarrow{i_3} G(X_2)\rightarrow G(X_2\setminus X_3)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_2)$ is generated by the classes $[\mathcal{O}_{X_2}]$ and $[\mathcal{O}_{X_3}]$. Similarly, from $$G(X_2)\xrightarrow{i_2} G(X_1)\rightarrow G(X_1\setminus X_2)\cong \mathbb{Z}\rightarrow 0,$$ we see that $G(X_1)$ is generated by $[\mathcal{O}_{X_1}], [\mathcal{O}_{X_2}]$, and $[\mathcal{O}_{X_3}]$. Lastly, $$G(X_1)\xrightarrow{i_1} G(X_0)\rightarrow G(X_0\setminus X_1)\cong \mathbb{Z}\rightarrow 0,$$ shows that $G(X_0)=G(X)$ is generated by the classes $\{[\mathcal{O}_{X_i}]\}_i$ where $i=0,1,2,3$ (in particular, $G(X)$ is a finitely generated abelian group).

Although it can be shown that $i_3$ and $i_2$ are split injections, I'm not immediately sure how to show $i_1$ is a split injection. (The morphism $i_3$ is split by the pushforward structure map $X_2\rightarrow \mathrm{Spec}(k)$, so this one is an injection, and $i_2$ is split by the pushforward along a contraction map $X_1\rightarrow X_2$.)

Instead, we can do this: let $F^i\subset G(X)$ denote the subgroup generated by classes $[\mathcal{O}_V]$ where $V$ is a subscheme of codimension $\mathrm{codim}_X(V)\geq i$. Then $F^0=G(X)$, and $F^0/F^1\cong \mathbb{Z}$ by the rank morphism; in particular, $G(X)\cong F^1\oplus \mathbb{Z}$ with $[\mathcal{O}_{X_0}]$ generating the second factor. The quotient $F^1/F^2$ is canonically the divisor class group $\mathrm{Cl}(X)$ which is isomorphic with $\mathbb{Z}\times \mathbb{Z}$ generated by a line in each fiber of the two projections $X\rightarrow \mathbb{P}^1$. This gives an isomorphism $$G(X)\cong F^1\oplus \mathbb{Z} \cong F^2\oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ Also, the map $G(X)\rightarrow F^1/F^2\cong \mathrm{Cl}(X)$ is defined so that $[\mathcal{O}_V]$ is sent to the class of the associated divisor $[V]$ for any integral subscheme $V\subset X$. Since there is an exact sequence (https://stacks.math.columbia.edu/tag/0C4J) $$0\rightarrow \mathcal{O}_{X_1}\rightarrow \mathcal{O}_{\mathbb{P}^1\times \infty}\oplus\mathcal{O}_{\infty \times \mathbb{P}^1}\rightarrow \mathcal{O}_{X_3}\rightarrow 0,$$ we see that $[\mathcal{O}_{X_1}]=[\mathcal{O}_{\mathbb{P}^1\times \infty}]+[\mathcal{O}_{\infty \times \mathbb{P}^1}] - [\mathcal{O}_{X_3}]$. It follows that $[\mathcal{O}_{X_1}]$ and $[\mathcal{O}_{X_2}]$ also generate $F^1/F^2$.

It's not too hard to see that $F^3=0$ and there is a map $F^2\rightarrow \mathbb{Z}$ gotten by pushing forward $F^2\subset G(X)\rightarrow G(\mathrm{Spec}(k))=\mathbb{Z}$, which is essentially just the degree map. The class $[\mathcal{O}_{X_3}]$ generates the image of this map and, if $K$ is the kernel $K=\mathrm{ker}(F^2\rightarrow \mathbb{Z})$ then $$G(X)=K\oplus \mathbb{Z} \oplus (\mathbb{Z}\times \mathbb{Z})\oplus \mathbb{Z}.$$ One can see $K=0$ a number of ways.

Edit: Here's another way that gets all the $G$-theory groups. Let $X_0=\mathbb{P}^1\times \mathbb{P}^1$ and $X_1=\mathbb{P}^1\times \infty$. Then $X_0\setminus X_1\cong \mathbb{P}^1\times \mathbb{A}^1$. Now the localization sequence looks like $$\cdots \rightarrow G_{i+1}(X_0\setminus X_1)\rightarrow G_i(X_1)\rightarrow G_i(X_0)\rightarrow G_i(X_0\setminus X_1)\rightarrow G_{i-1}(X_1)\rightarrow \cdots.$$

The map $G_i(X_1)\rightarrow G_i(X_0)$ is split by the pushforward along the first projection $X_0\rightarrow \mathbb{P}^1$. This gives a decomposition $G_i(X_0)=G_i(\mathbb{P}^1)\oplus G_i(\mathbb{P}^1\times \mathbb{A}^1)$. And by homotopy invariance, which was implicit in the above argument, there is an isomorphism $G_i(\mathbb{P}^1\times \mathbb{A}^1)\cong G_i(\mathbb{P}^1)$.