Here is a more precise formulation of my suggestion above. First of all, I do not believe that you can find such a surface in $\mathbb{P}^3$. The base locus curves must be coordinate lines. In order to get the behavior you describe, you will need the entire coordinate tetrahedron to be contained in the surface. But then the vertices where three coordinate lines intersect would be singular points of your surface. Thus, the first place to look is in $\mathbb{P}^4$.

Let $[x_0,x_1,x_2,x_3,x_4]$ be the standard homogeneous coordinates on $\mathbb{P}^4$. For every $0\leq i < j \leq 4$, denote by $\Pi_{\{i,j\} }$ the common zero locus $\text{Zero}(x_i,x_j)$. This is a $2$-plane. For every $0\leq i < j < k \leq 4$, denote by $L_{\{i,j,k\} }$ the common zero locus $\text{Zero}(x_i,x_j,x_k)$. This is a line. Inside each $\Pi_{\{i,j\} }$, let $B_{\{i,j\} }$ be a line distinct from each $L_{\{i,j,k\} }$ and such that the points of intersection $p_{\{i,j\};k}\in L_{\{i,j,k\} }\cap B_{\{i,j\} }$ satisfy $(p_{\{i,j\} ;k},p_{\{j,k\} ;i},p_{\{k,i\} ;j})$ are three distinct points of $L_{\{i,j,k\} }$. Let $B$ be the curve $\cup_{i,j} B_{\{i,j\} }$. This is a disjoint union of $10$ lines.

Denote by $\mathcal{I}_{B/\mathbb{P}^4}$ the ideal sheaf of $B$ in $\mathbb{P}^4$. By Serre vanishing, there exists an integer $d_0$ (which we could bound above, if it were important) such that for every integer $d\geq d_0$, both $h^1(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d))$ and $h^1(\mathbb{P}^4,\mathcal{I}^2_{B/\mathbb{P}^4}(d))$ equal zero. Thus, the following restriction map is surjective, $$r_d:V_d\to W_d,$$ $$H^0(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d)) \to H^0(\mathbb{P}^4,\left( \mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4} \right)(d)). $$ Of course $\mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4}$ is the pushforward from $B$ of a locally free sheaf of rank $3$. For every line $M = B_{\{ i,j \} }$, the locally free sheaf on $M$ is $N_{M/\mathbb{P}^4}^{\vee} \cong \mathcal{O}_M(-1)^{\oplus 3}$, the conormal sheaf of $M$ in $\mathbb{P}^4$. Thus, each section of $W_d$ can be interpreted as a collection of sheaf homomorphisms $N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M$, one for each of the $10$ lines in $B$.

For integers $d,e\geq d_0$, a pair $(s,t) \in W_d\times W_e$ can be interpreted as a collection of sheaf homomorphisms $$\phi_{s,t,N}:N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M \oplus \mathcal{O}(e)|_M,$$ one for each of the $10$ lines in $B$. The surjectivity above implies that every such $10$-tuple of sheaf homomorphisms arises from a pair in $V_d\times V_e$. For a general such sheaf homomorphism $\phi$ on $M$, the cokernel is locally free of rank $1$. For instance, if we choose homogeneous coordinates $y_0,y_1$ on $M$, then for $d,e\geq 1$, the following homomorphism has this property, $$\phi:\mathcal{O}(1)^{\oplus 3} \to \mathcal{O}(d)\oplus \mathcal{O}(e), \ \ \ \left[ \begin{array}{rrr} y_0^{d-1} & y_1^{d-1} & 0 \\ 0 & y_0^{e-1} & y_1^{e-1} \end{array} \right] .$$

Because of this, for a general choice $(\widetilde{s},\widetilde{t})$ in $V_d\times V_e$, interpreting $\widetilde{s}$ and $\widetilde{t}$ as global sections of $\mathcal{O}_{\mathbb{P}^4}(d)$, resp. $\mathcal{O}_{\mathbb{P}^4}(e)$, the common zero locus $X$ of $\widetilde{s}$ and $\widetilde{t}$ contains $B$, and is everywhere a smooth surface along $B$. By Bertini's theorem, $X$ is also everywhere a smooth surface away from the base locus $B$. Thus, $X$ is a smooth complete intersection surface in $\mathbb{P}^4$ that contains $B$. In particular, because $h^1$ and $h^2$ of invertible sheaves on $\mathbb{P}^4$ are zero, it follows that the following restriction map is an isomorphism, $$H^0(\mathbb{P}^4,\mathcal{O}(1)) \to H^0(X,\mathcal{O}(1)|_X).$$ Thus, the standard basis $x_0,\dots,x_4$ for $H^0(\mathbb{P}^4,\mathcal{O}(1))$ maps to a basis for $H^0(X,\mathcal{O}(1)|_X)$. By construction, for every $0\leq i<j \leq 4$, the intersection of $X$ with the common zero locus $\text{Zero}(x_i,x_j)$ contains a line $B_{\{i,j\} }$.

Here is a more precise formulation of my suggestion above. First of all, I do not believe that you can find such a surface in $\mathbb{P}^3$. The base locus curves must be coordinate lines. In order to get the behavior you describe, you will need the entire coordinate tetrahedron to be contained in the surface. But then the vertices where three coordinate lines intersect would be singular points of your surface. Thus, the first place to look is in $\mathbb{P}^4$.

Let $[x_0,x_1,x_2,x_3,x_4]$ be the standard homogeneous coordinates on $\mathbb{P}^4$. For every $0\leq i < j \leq 4$, denote by $\Pi_{\{i,j\} }$ the common zero locus $\text{Zero}(x_i,x_j)$. This is a $2$-plane. For every $0\leq i < j < k \leq 4$, denote by $L_{\{i,j,k\} }$ the common zero locus $\text{Zero}(x_i,x_j,x_k)$. This is a line. Inside each $\Pi_{\{i,j\} }$, let $B_{\{i,j\} }$ be a line distinct from each $L_{\{i,j,k\} }$ and such that the points of intersection $p_{\{i,j\};k}\in L_{\{i,j,k\} }\cap B_{\{i,j\} }$ satisfy $(p_{\{i,j\} ;k},p_{\{j,k\} ;i},p_{\{k,i\} ;j})$ are three distinct points of $L_{\{i,j,k\} }$. Let $B$ be the curve $\cup_{i,j} B_{\{i,j\} }$. This is a disjoint union of $10$ lines.

Denote by $\mathcal{I}_{B/\mathbb{P}^4}$ the ideal sheaf of $B$ in $\mathbb{P}^4$. By Serre vanishing, there exists an integer $d_0$ (which we could bound above, if it were important) such that for every integer $d\geq d_0$, both $h^1(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d))$ and $h^1(\mathbb{P}^4,\mathcal{I}^2_{B/\mathbb{P}^4}(d))$ equal zero. Thus, the following restriction map is surjective, $$r_d:V_d\to W_d,$$ $$H^0(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d)) \to H^0(\mathbb{P}^4,\left( \mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4} \right)(d)). $$ Of course $\mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4}$ is the pushforward from $B$ of a locally free sheaf of rank $3$. For every line $M = B_{\{ i,j \} }$, the locally free sheaf on $M$ is $N_{M/\mathbb{P}^4}^{\vee} \cong \mathcal{O}_M(-1)^{\oplus 3}$, the conormal sheaf of $M$ in $\mathbb{P}^4$. Thus, each section of $W_d$ can be interpreted as a collection of sheaf homomorphisms $N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M$, one for each of the $10$ lines in $B$.

For integers $d,e\geq d_0$, a pair $(s,t) \in W_d\times W_e$ can be interpreted as a collection of sheaf homomorphisms $$\phi_{s,t,N}:N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M \oplus \mathcal{O}(e)|_M,$$ one for each of the $10$ lines in $B$. The surjectivity above implies that every such $10$-tuple of sheaf homomorphisms arises from a pair in $V_d\times V_e$. For a general such sheaf homomorphism $\phi$ on $M$, the cokernel kernel is locally free of rank $1$ and the homormophism is surjective. For instance, if we choose homogeneous coordinates $y_0,y_1$ on $M$, then for $d,e\geq 1$, the following homomorphism has this property, $$\phi:\mathcal{O}(1)^{\oplus 3} \to \mathcal{O}(d)\oplus \mathcal{O}(e), \ \ \ \left[ \begin{array}{rrr} y_0^{d-1} & y_1^{d-1} & 0 \\ 0 & y_0^{e-1} & y_1^{e-1} \end{array} \right] .$$

Because of this, for a general choice $(\widetilde{s},\widetilde{t})$ in $V_d\times V_e$, interpreting $\widetilde{s}$ and $\widetilde{t}$ as global sections of $\mathcal{O}_{\mathbb{P}^4}(d)$, resp. $\mathcal{O}_{\mathbb{P}^4}(e)$, the common zero locus $X$ of $\widetilde{s}$ and $\widetilde{t}$ contains $B$, and is everywhere a smooth surface along $B$. By Bertini's theorem, $X$ is also everywhere a smooth surface away from the base locus $B$. Thus, $X$ is a smooth complete intersection surface in $\mathbb{P}^4$ that contains $B$. In particular, because $h^1$ and $h^2$ of invertible sheaves on $\mathbb{P}^4$ are zero, it follows that the following restriction map is an isomorphism, $$H^0(\mathbb{P}^4,\mathcal{O}(1)) \to H^0(X,\mathcal{O}(1)|_X).$$ Thus, the standard basis $x_0,\dots,x_4$ for $H^0(\mathbb{P}^4,\mathcal{O}(1))$ maps to a basis for $H^0(X,\mathcal{O}(1)|_X)$. By construction, for every $0\leq i<j \leq 4$, the intersection of $X$ with the common zero locus $\text{Zero}(x_i,x_j)$ contains a line $B_{\{i,j\} }$.

Here is a more precise formulation of my suggestion above. First of all, I do not believe that you can find such a surface in $\mathbb{P}^3$. The base locus curves must be coordinate lines. In order to get the behavior you describe, you will need the entire coordinate tetrahedron to be contained in the surface. But then the vertices where three coordinate lines intersect would be singular points of your surface. Thus, the first place to look is in $\mathbb{P}^4$.

Let $[x_0,x_1,x_2,x_3,x_4]$ be the standard homogeneous coordinates on $\mathbb{P}^4$. For every $0\leq i < j \leq 4$, denote by $\Pi_{\{i,j\} }$ the common zero locus $\text{Zero}(x_i,x_j)$. This is a $2$-plane. For every $0\leq i < j < k \leq 4$, denote by $L_{\{i,j,k\} }$ the common zero locus $\text{Zero}(x_i,x_j,x_k)$. This is a line. Inside each $\Pi_{\{i,j\} }$, let $B_{\{i,j\} }$ be a line distinct from each $L_{\{i,j,k\} }$ and such that the points of intersection $p_{\{i,j\};k}\in L_{\{i,j,k\} }\cap B_{\{i,j\} }$ satisfy $(p_{\{i,j\} ;k},p_{\{j,k\} ;i},p_{\{k,i\} ;j})$ are three distinct points of $L_{\{i,j,k\} }$. Let $B$ be the curve $\cup_{i,j} B_{\{i,j\} }$. This is a disjoint union of $10$ lines.

Denote by $\mathcal{I}_{B/\mathbb{P}^4}$ the ideal sheaf of $B$ in $\mathbb{P}^4$. By Serre vanishing, there exists an integer $d_0$ (which we could bound above, if it were important) such that for every integer $d\geq d_0$, both $h^1(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d))$ and $h^1(\mathbb{P}^4,\mathcal{I}^2_{B/\mathbb{P}^4}(d))$ equal zero. Thus, the following restriction map is surjective, $$r_d:V_d\to W_d,$$ $$H^0(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d)) \to H^0(\mathbb{P}^4,\left( \mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4} \right)(d)). $$ Of course $\mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4}$ is the pushforward from $B$ of a locally free sheaf of rank $3$. For every line $M = B_{\{ i,j \} }$, the locally free sheaf on $M$ is $N_{M/\mathbb{P}^4}^{\vee} \cong \mathcal{O}_M(-1)^{\oplus 3}$, the conormal sheaf of $M$ in $\mathbb{P}^4$. Thus, each section of $W_d$ can be interpreted as a collection of sheaf homomorphisms $N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M$, one for each of the $10$ lines in $B$.

For integers $d,e\geq d_0$, a pair $(s,t) \in W_d\times W_e$ can be interpreted as a collection of sheaf homomorphisms $$\phi_{s,t,N}:N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M \oplus \mathcal{O}(e)|_M,$$ one for each of the $10$ lines in $B$. For a general such choice of $s$ and $t$, each $\phi_{s,t,N}$ has cokernel that is locally free of rank $1$. For instance, if we choose homogeneous coordinates $y_0,y_1$ on $M$, then for $d,e\geq 1$, then the following homomorphism has this property, $$\phi:\mathcal{O}(1)^{\oplus 3} \to \mathcal{O}(d)\oplus \mathcal{O}(e), \ \ \ \left[ \begin{array}{rrr} y_0^{d-1} & y_1^{d-1} & 0 \\ 0 & y_0^{e-1} & y_1^{e-1} \end{array} \right] .$$

Because of this, for a general choice $(s,t)$ in $U_d\times U_e$, interpreting $s$ and $t$ as global sections of $\mathcal{O}_{\mathbb{P}^4}(d)$, resp. $\mathcal{O}_{\mathbb{P}^4}(e)$, the common zero locus $X$ of $s$ and $t$ is everywhere smooth along $B$. By Bertini's theorem, $X$ is also everywhere smooth away from the base locus $B$. Thus, $X$ is a smooth complete intersection surface in $\mathbb{P}^4$. In particular, because $h^1$ and $h^2$ of invertible sheaves on $\mathbb{P}^4$ are zero, it follows that the following restriction map is an isomorphism, $$H^0(\mathbb{P}^4,\mathcal{O}(1)) \to H^0(X,\mathcal{O}(1)|_X).$$ Thus, the standard basis $x_0,\dots,x_4$ for $H^0(\mathbb{P}^4,\mathcal{O}(1))$ maps to a basis for $H^0(X,\mathcal{O}(1)|_X)$. By construction, for every $0\leq i<j \leq 4$, the intersection of $X$ with the common zero locus $\text{Zero}(x_i,x_j)$ contains a line $B_{\{i,j\} }$.

Here is a more precise formulation of my suggestion above. First of all, I do not believe that you can find such a surface in $\mathbb{P}^3$. The base locus curves must be coordinate lines. In order to get the behavior you describe, you will need the entire coordinate tetrahedron to be contained in the surface. But then the vertices where three coordinate lines intersect would be singular points of your surface. Thus, the first place to look is in $\mathbb{P}^4$.

Let $[x_0,x_1,x_2,x_3,x_4]$ be the standard homogeneous coordinates on $\mathbb{P}^4$. For every $0\leq i < j \leq 4$, denote by $\Pi_{\{i,j\} }$ the common zero locus $\text{Zero}(x_i,x_j)$. This is a $2$-plane. For every $0\leq i < j < k \leq 4$, denote by $L_{\{i,j,k\} }$ the common zero locus $\text{Zero}(x_i,x_j,x_k)$. This is a line. Inside each $\Pi_{\{i,j\} }$, let $B_{\{i,j\} }$ be a line distinct from each $L_{\{i,j,k\} }$ and such that the points of intersection $p_{\{i,j\};k}\in L_{\{i,j,k\} }\cap B_{\{i,j\} }$ satisfy $(p_{\{i,j\} ;k},p_{\{j,k\} ;i},p_{\{k,i\} ;j})$ are three distinct points of $L_{\{i,j,k\} }$. Let $B$ be the curve $\cup_{i,j} B_{\{i,j\} }$. This is a disjoint union of $10$ lines.

Denote by $\mathcal{I}_{B/\mathbb{P}^4}$ the ideal sheaf of $B$ in $\mathbb{P}^4$. By Serre vanishing, there exists an integer $d_0$ (which we could bound above, if it were important) such that for every integer $d\geq d_0$, both $h^1(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d))$ and $h^1(\mathbb{P}^4,\mathcal{I}^2_{B/\mathbb{P}^4}(d))$ equal zero. Thus, the following restriction map is surjective, $$r_d:V_d\to W_d,$$ $$H^0(\mathbb{P}^4,\mathcal{I}_{B/\mathbb{P}^4}(d)) \to H^0(\mathbb{P}^4,\left( \mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4} \right)(d)). $$ Of course $\mathcal{I}_{B/\mathbb{P}^4}/\mathcal{I}^2_{B/\mathbb{P}^4}$ is the pushforward from $B$ of a locally free sheaf of rank $3$. For every line $M = B_{\{ i,j \} }$, the locally free sheaf on $M$ is $N_{M/\mathbb{P}^4}^{\vee} \cong \mathcal{O}_M(-1)^{\oplus 3}$, the conormal sheaf of $M$ in $\mathbb{P}^4$. Thus, each section of $W_d$ can be interpreted as a collection of sheaf homomorphisms $N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M$, one for each of the $10$ lines in $B$.

For integers $d,e\geq d_0$, a pair $(s,t) \in W_d\times W_e$ can be interpreted as a collection of sheaf homomorphisms $$\phi_{s,t,N}:N_{M/\mathbb{P}^4}\to \mathcal{O}(d)|_M \oplus \mathcal{O}(e)|_M,$$ one for each of the $10$ lines in $B$. The surjectivity above implies that every such $10$-tuple of sheaf homomorphisms arises from a pair in $V_d\times V_e$. For a general such sheaf homomorphism $\phi$ on $M$, the cokernel is locally free of rank $1$. For instance, if we choose homogeneous coordinates $y_0,y_1$ on $M$, then for $d,e\geq 1$, the following homomorphism has this property, $$\phi:\mathcal{O}(1)^{\oplus 3} \to \mathcal{O}(d)\oplus \mathcal{O}(e), \ \ \ \left[ \begin{array}{rrr} y_0^{d-1} & y_1^{d-1} & 0 \\ 0 & y_0^{e-1} & y_1^{e-1} \end{array} \right] .$$

Because of this, for a general choice $(\widetilde{s},\widetilde{t})$ in $V_d\times V_e$, interpreting $\widetilde{s}$ and $\widetilde{t}$ as global sections of $\mathcal{O}_{\mathbb{P}^4}(d)$, resp. $\mathcal{O}_{\mathbb{P}^4}(e)$, the common zero locus $X$ of $\widetilde{s}$ and $\widetilde{t}$ contains $B$, and is everywhere a smooth surface along $B$. By Bertini's theorem, $X$ is also everywhere a smooth surface away from the base locus $B$. Thus, $X$ is a smooth complete intersection surface in $\mathbb{P}^4$ that contains $B$. In particular, because $h^1$ and $h^2$ of invertible sheaves on $\mathbb{P}^4$ are zero, it follows that the following restriction map is an isomorphism, $$H^0(\mathbb{P}^4,\mathcal{O}(1)) \to H^0(X,\mathcal{O}(1)|_X).$$ Thus, the standard basis $x_0,\dots,x_4$ for $H^0(\mathbb{P}^4,\mathcal{O}(1))$ maps to a basis for $H^0(X,\mathcal{O}(1)|_X)$. By construction, for every $0\leq i<j \leq 4$, the intersection of $X$ with the common zero locus $\text{Zero}(x_i,x_j)$ contains a line $B_{\{i,j\} }$.