It may help to be familiar with the concept of scattering parameters or S-parameters. The S-parameters are the elements of a matrix which tell you the amplitude and phase coming out of every port, for the signal going into any other port. It assumes all the ports are terminated in matched loads, and it's usually valid for only one frequency.
For a 3-port device, there are 9 S-parameters. They are written in a matrix like this:
$$
[S] =
\begin{bmatrix}
S_{11} & S_{12} & S_{13} \\
S_{21} & S_{22} & S_{23} \\
S_{31} & S_{32} & S_{33}
\end{bmatrix}
$$
The subscripts are ordered output, then input. So, $S_{21}$ is related to how the input at port 1 contributes to what's observed at port 2.
In the case of feeding an array of two antennas, you'll probably use a Wilkinson power divider. Such a power divider with a 50/50 power split, where port 1 is the transceiver and ports 2 and 3 are the antennas, has a scattering matrix of:
$$
[S] =
\begin{bmatrix}
0 & {-j \over \sqrt 2} & {-j \over \sqrt 2} \\
{-j \over \sqrt 2} & 0 & 0 \\
{-j \over \sqrt 2} & 0 & 0
\end{bmatrix}
$$
Consider transmitting: we're really only concerned with the power in to port 1, so we only need to look at the first column of the matrix. $S_{11}$ is 0, meaning none of the power into port 1 is reflected back out port 1, meaning the SWR is 1:1.
$S_{21}$ and $S_{31}$ are each ${-j \over \sqrt 2}$. The $-j$ term just means the phase of the output is rotated 90 degrees from the input, which would make sense because the power divider is constructed from 1/4-wave transmission lines. The need for the $\sqrt{2}$ term should be obvious considering this is a passive network, and here we're also assuming ideal, lossless components. Thus, power out can't be more than power in. The S-parameters relate to amplitude and phase, so we must square them to get power. $ \left(1/\sqrt 2\right)^2 = {1 / 2} $, so half the power coming in port 1 goes out port 2, the other half goes out port 3.
Now what if a signal is applied to port 2? $S_{12} = {-j \over \sqrt 2}$, so half the power goes back to the transceiver. $S_{22} = S_{32} = 0$, meaning none of the power goes to the other ports. The remaining half of the power is lost in the resistor.
Port 3 is identical to port 2.
However as a receiver, if a linearly polarized signal is received, it may be picked up by only one of the two antennas, and the matching network will deliver some power to the feed line (as desired), but also some to the second (90 deg) antenna, which will re-radiate it thus degrading the overall antenna gain.
Not quite! Because $S_{23} = S_{32} = 0$, none of the power received at one antenna port makes it to the other antenna port. There is no re-radiation. But because $S_{12} = S_{13} = {-j \over \sqrt 2}$, only half the power received at one of the antennas (assuming for now that only one antenna is receiving a signal) makes it to the receiver, which does indeed reduce antenna gain by 3 dB.
Now what if both antennas are receiving a signal simultaneously? Let's say the signal at each antenna port is $1 / \sqrt 2$, so that combined they have power 1. In order to determine what appears at the receiver, we must calculate:
$$ \begin{align}
&S_{12} {1 \over \sqrt 2} + S_{13} {1 \over \sqrt 2} \\
= &{-j \over \sqrt 2} {1 \over \sqrt 2} + {-j \over \sqrt 2} {1 \over \sqrt 2} \\
= &{-j \over 2} + {-j \over 2}\\
= &-j
\end{align}$$
$-j$ is just 1 with a 90 degree phase shift, so all the power received by the antennas makes it to the receiver, just shifted in phase.
But this only works when the signals are in phase. What if port 2 is receiving $1 / \sqrt 2$ but port 3 is receiving the same thing, but 90 degrees shifted: $-j / \sqrt 2 $?
$$ \begin{align}
&S_{12} {1 \over \sqrt 2} + S_{13} {-j \over \sqrt 2} \\
= &{-j \over \sqrt 2} {1 \over \sqrt 2} + {-j \over \sqrt 2} {-j \over \sqrt 2} \\
= &{-j \over 2} + {1 \over 2}
\end{align}$$
Converting this to polar form we find the magnitude is $1/\sqrt 2$, so we lost half the power. So in the case of receiving, the Wilkinson power divider (or I guess combiner, in this direction) is lossless only when ports 2 and 3 are in phase.
Unsurprisingly, repeating the same thing with the antennas 180 degrees out of phase results in all the power lost.
When you want to generate circular polarization, the transmitter will feed the power divider, then one of the power divider's outputs will have an additional 90 degree delay in it. On receive, this delay has the effect of combining received signals of the same circular polarization to be in-phase at the power divider's (or now, combiner) antenna ports to be in-phase, and they will make it to the receiver without loss.
If the received signal is circularly polarized with opposite chirality, the signals are opposite phase at the power combiner, and the loss is theoretically infinite.
If the received signal is linearly polarized (at any angle, horizontal, vertical, or otherwise) then the phase shift is 90 degrees and half the power is lost.
There is no way around this with just receiver and one antenna. More generally, any antenna or antenna array you build must pick one point on the Poincaré sphere, and there will always be an opposite point on that sphere that's orthogonal to your antenna which you can't receive, with all other possible polarizations on the sphere having some loss between zero and infinity.
If you want to receive any polarization without loss, then you must build two antennas and two receivers, each occupying opposite points on the Poincaré sphere. There is then some combination of amplitude and phase coefficients with which those signals can be combined to receive an incoming signal at any polarization with zero loss.
