Fix a curve $C$ and a line both in $\mathbb{P}^3$ not intersecting one another. Suppose that the ideal of the curve is generated by $P_i$ and the line is definied by linear polynomials $l_1, l_2$. Is it true that a general degree $d$ smooth surface in $\mathbb{P}^3$ containing $C \cup l$ is defined by an equation of the form $$ l_1\left(\sum_iP_iQ_i\right)+l_2\left(\sum_iP_iQ'_i\right) $$ for $Q_i, Q'_i$ homogeneous polynomials of degree equal to $d1\deg(P_i)$?
The answer is yes if and only if $H^1(J(k))=0$ for all $k$, where $J$ defines $C$. It can be seen as follows. Let $I$ be the ideal defining $l$. Then we have the Koszul resolution, $0\to\mathcal{O}(2)\to\mathcal{O}(1)^2\to I\to 0$. This gives a complex, $0\to J(2)\to J(1)^2\to JI\to 0$ and since the curves are disjoint, $JI=J\cap I$. One easily checks that this is in fact exact. If $H^1(J(k)=0$ for all $k$, we get a surjection $H^0(J(1+k))^2\to H^0(J\cap I(k))$ and this gives what you want. If on the other hand, $H^1(J(k))\neq 0$ for some $k$, choose $k$ the largest such. Then, $H^0(J(k+1))^2\to H^0(J\cap I(k+2))$ is not surjective. 


The answer is no in general. Any surface of degree $d$ passing through the lines is of the form $l_1 Q_1+l_2Q_2$ where $Q_1,Q_2$ are homogeneous polynomials of degree $d1$. Since the line does not intersect the curve $C$, the curve $C$ is not in a plane passing through the line. Hence, neither $l_1$ neither $l_2$ vanishes on $C$. But this does not imply that $Q_1,Q_2$ vanish on $C$. Let us give a precise example (sorry my previous one did not work because $C$ and $l$ were not empty, but this one is OK). It suffices to choose a smooth cubic $S$ in $\mathbb{P}^3$, a line $l$ on it, and a curve $C$ on $S$ which does not intersect $l$ and which is not contained in any quadric. Then, the equation of $S$ if of the form $l_1Q_1+l_2Q_2$ where $Q_1,Q_2$ have degree $2$. BUT it is not possible to choose $Q_1,Q_2$ that vanish on $C$, since $C$ is not contained in any quadric. It remains to see that such a curve $C$ exists; and in fact a general curve of $S$ will make the job. There exists a birational morphism $S\to \mathbb{P}^2$ which contracts $6$ lines of $S$ onto $6$ points of $\mathbb{P}^2$ in general position, and we can choose that $l$ is one of the lines contracted. Take then a curve of $\mathbb{P}^2$ of high degree (say $d>2$) not passing through any of the $6$ points. The preimage on $S$ gives a curve $C$ on $S$ which does not intersect $l$. Moreover, the degree of $C$ in $\mathbb{P}^3$ is the intersection with a hyperplane section, corresponding to a cubic of $\mathbb{P}^2$ though the six points, and is thus $3d>6$. Since $C$ is contained in a smooth cubic, and has degree $>6$, it is not contained in a quadric. PS: The same argument does not work in smaller degree, i.e. if $S$ is a quadric. Indeed, the only curves in $S$ which do not intersect $l$ are lines, and are then contained in planes. 


Consider the ideal generated by $(l_1 P_i, l_2P_i)=(l_1,l_2)(P_i)$. You are asking whether a generic function vanishing on the line and the curve is contained in this ideal. Clearly the vanishing subscheme of this ideal is just the union of the curve and the line  we can check this by working locally. But homogeneous ideals are not quite the same thing as sheaves of ideals on projective space. $I$ and $(x_0I,x_1I,x_2I,x_3I)$ will always have the same vanishing set. However, this is the only thing that can happen  if $f$ vanishes on $V(I)$, then $f$ is in $I$ on each open affine, so $x_i^n f$ is in $I$ for any $i$ and for sufficiently large $n$. It's not immediately obvious to me whether this can happen in your situation. But it clearly can only happen for polynomials of very small degree. 

