I know that field lines are directly proportional to the value of electric charge. Assuming that $8$ lines of force, or of field, of the come out of the charge $+q$, then $16$ lines of force will come out of the charge $+2q$ since from the electrostatic field flux $\Phi_S(E)$ we have:
$$\Phi_S(E)\propto (+q)$$
where $S$ it is the Gaussian surface.
In a Physics textbook for students of a high school that does not introduce the flow of an electrostatic field and the subsequent concepts, there are some pictures of an electric dipole, a system formed by two equal and opposite charges, $+q$ and $-q$, separated by a non-zero distance. It is observed that if the algebraic sum of the charges is zero, and part of the lines of force extend to infinity and part, clearly, are the lines of force that start from the positive charge and close on the negative charge.
In the case of two positive and negative charges, the drawing that represented the lines of force is clear, but I observed that the algebraic sum of the charges is not zero.
Is there a rigorous mathematical proof of the reason that two charges of sign $+2q$ and $-q$, although not having algebraic sum zero, do not have field lines extending infinitely from charge $-q$ but the field lines are all closed relative to charge $-q$ starting from $+2q$? By drawing the electrostatic field with a test charge $+q_0$, the field lines are all close in $-q$.
Best Answer
Is there a rigorous mathematical proof of the reason that . . . . .
I cannot give you a rigorous mathematical proof but can give you an indication of what the electric field looks like far away from the charges.
The electric field would be practically indistinguishable from the the electric field due to a single charge $+q$.
The electric field lines would be radial and pointing outwards from the vicinity of the two charges.
The electric field diagram is misleading it does not show that there is a neutral (zero field) point $(1+\sqrt 2)d$ to the right of the $-q$ charge. $d$ is the separation of the charges.
Beyond the neutral point the electric field line is pointing away from the charges (to the right).
To find where the neutral point, $N$, is consider the following diagram.
At the neutral point the electric fields due to the two charges are equal in magnitude and opposite in direction.
$\vec E_{\rm +2q} + \vec E_{\rm -q} = \vec 0 \Rightarrow k\dfrac {+2q}{(d+x)^2} \hat i + k\dfrac {-q}{x^2} \hat i = \vec 0 \Rightarrow \dfrac {2}{(d+x)^2} = \dfrac {1}{x^2} \Rightarrow x = (1+\sqrt 2)d$
I have produced a better set of diagrams to illustrates the points that I have made in my answer. Note the charge of scale from diagram to diagram.