Your intuition is correct $-$ an asymmetry between $\alpha$ and $\beta$ is only possible if the two charges are different.
If you want a quantitative relationship between the two angles, the correct (read: the only viable) approach is via Gauss's law for the electric flux. From the geometry of the field, which is symmetric about the inter-charge axis, it is relatively easy to see that if you take the surface of revolution generated by the field line about the axis, then you get a cylinder-with-two-conical-ends which goes from one charge to the other:
By definition, the electric field is tangential to this surface at every point, which means that no field lines leave it, and the electric flux is confined inside it. That means, therefore, that the electric flux that leaves charge 1 into the surface must equal the electric flux that arrives at charge 2.
Moreover, we know how to relate these electric fluxes to the angles: close to charge 1, we can ignore the effect of charge 2, and then we just have the flat integral, i.e., the electric flux is the product of the charge times the solid angle spanned by the cone at its apex,
$$
\Phi_1 = q_1 \: \Omega_1,
$$
where the solid angle can be calculated explicitly as
\begin{align}
\Omega_1
& = \int_0^\alpha \int_0^2\pi \sin(\theta)\mathrm d\phi \:\mathrm d\theta
\\ & = 2\pi \big[-\cos(\theta)\big]_0^\alpha
\\ & = 2\pi(1-\cos(\alpha)).
\end{align}
Assuming that $q_1>0>q_2$, we can set $\Phi_1+\Phi_2=0$ and therefore
\begin{align}
2\pi(1-\cos(\alpha))|q_1| & = 2\pi(1-\cos(\beta))|q_2|
\\ \implies
\frac{1-\cos(\alpha)}{1-\cos(\beta)} & = \frac{|q_2|}{|q_1|}
.
\end{align}
This relationship then allows you to find any of the relevant quantities in terms of the other three, which is the most that you can do here.
It is correct that $s = r \sin \alpha$, but since $r$ depends on $\alpha$ it does not follow that $ds = r \cos \alpha$. Instead , you have $ds = \left ( r \cos \alpha + dr/d\alpha \sin \alpha \right) d\alpha $. If you work that out you will recover the textbook substitution.
Best Answer
I think this solution will answer all of your questions. Note here that $k=1/(4\pi\epsilon_0)$. The electric field $\vec{E}$ for any given charge density distribution $\rho(\vec{r}')$ is
\begin{align} \vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int \rho(\vec{r}')\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}d^3r'\,, \end{align}
where $\vec{r}$ is the vector pointing from the origin to the point at which the field is to be calculated (in your case, pointing to point $P$) and $\vec{r}'$ is the vector pointing from the origin to the distribution of charge, which will be integrated over. Placing the origin of the cylindrical coordinate system $(r,\phi,z)$ on the line of charge directly to the left of point $P$, then point $P$ is at $\vec{r}=r\,\hat{r}$. The line charge runs along the $z$-axis such that a general point on the line charge is denoted by $\vec{r}'=z\,\hat{z}$. The origin is intentionally placed such that $\vec{r}\perp\vec{r}'$, which will be very useful.
Since this is a line charge with linear charge density $\lambda$, then the differential charge volume element $dq=\rho(\vec{r}')\,d^3r'$ reduces to $dq=\lambda\,dz$. The electric field now is:
\begin{align} \vec{E}(r) = \frac{\lambda}{4\pi\epsilon_0}\int_{-a}^b \frac{r\,\hat{r}-z\,\hat{z}}{|r\hat{r}-z\hat{z}|^3}dz\,. \end{align}
Here $-a$ and $b$ are the endpoints of the line charge on the $z$-axis, which can be taken to infinity later if desired. Now, recall that $\vec{r}\perp\vec{r}'$. This means that a right-triangle has been formed between point $P$ at $\vec{r}=r\,\hat{r}$, the origin, and the general point $\vec{r}'=z\,\hat{z}$ on the line charge. This means that the distance been $\vec{r}$ and $\vec{r}'$ (that is, the hypotenuse of the right triangle) is given by:
\begin{align} |r\,\hat{r}-z\,\hat{z}| = (r^2 + z^2)^{1/2}\,. \end{align}
Now the integral becomes
\begin{align} \vec{E}(r) = \frac{\lambda}{4\pi\epsilon_0}\int_{-a}^b \frac{r\,\hat{r}-z\,\hat{z}}{(r^2 + z^2)^{3/2}}dz\,. \end{align}
Here we can define the angle $\theta$ in the right-triangle such that $\tan{\theta}=\frac{z}{r}$, which allows us to make the trig substitution $z=r\tan{\theta}$, where $dz=r\sec^2{\theta}\,d\theta$. The integral now becomes
\begin{align} \vec{E}(r) &= \frac{\lambda}{4\pi\epsilon_0}\int_{-\alpha}^\beta \frac{r\,\hat{r}-r\tan{\theta}\,\hat{z}}{(r^2 + r^2\tan^2{\theta})^{3/2}}(r\sec^2{\theta}\,d\theta) \\ &= \frac{\lambda}{4\pi\epsilon_0 r}\int_{-\alpha}^\beta \frac{\hat{r}-\tan{\theta}\,\hat{z}}{(1 + \tan^2{\theta})^{3/2}}\sec^2{\theta}\,d\theta\,. \end{align}
Using the trig identity $1 + \tan^2{\theta}=\sec^2{\theta}$, the integral reduces to
\begin{align} \vec{E}(r) &= \frac{\lambda}{4\pi\epsilon_0 r}\int_{-\alpha}^\beta \frac{\hat{r}-\tan{\theta}\,\hat{z}}{\sec{\theta}}d\theta \\ &= \frac{\lambda}{4\pi\epsilon_0 r}\int_{-\alpha}^\beta \left[\cos{\theta}\,\hat{r}-\sin{\theta}\,\hat{z}\right]d\theta \\ &= \frac{\lambda}{4\pi\epsilon_0 r} \left[\sin{\theta}\,\hat{r}+\cos{\theta}\,\hat{z} \right]_{-\alpha}^\beta\,, \end{align}
where $\alpha=\arctan{\left(\frac{a}{r}\right)}$ and $\beta=\arctan{\left(\frac{b}{r}\right)}$. At this point, you can either keep the integral in terms of $\theta$ and evaluate it at $\alpha$ and $\beta$, or switch it back to the original variable $z$. If you choose to switch, one obtains:
\begin{align} \vec{E}(r) &= \frac{\lambda}{4\pi\epsilon_0 r} \left[\frac{z\,\hat{r}}{(r^2+z^2)^{1/2}}+\frac{r\,\hat{z}}{(r^2+z^2)^{1/2}} \right]_{-a}^b\,, \end{align}
given that $\sin{\theta}=\frac{z}{(r^2+z^2)^{1/2}}$ and $\cos{\theta}=\frac{r}{(r^2+z^2)^{1/2}}$. Either way, when taken to infinity the integral gives the desired result:
\begin{align} \vec{E}(r) = \frac{\lambda\,\hat{r}}{4\pi\epsilon_0 r} \left[1+1 \right] = \frac{\lambda\,\hat{r}}{2\pi\epsilon_0 r} = \frac{2k\lambda}{r}\hat{r}\,. \end{align}
Hope this helps.