Can the round function $F$ in a Feistel Network be practically any non-invertible function? This question might seem silly, but how true is the fact that the round function $F$ does not have to be invertible?

I am curious to know this, because non-invertible functions can be very lossy, i.e., can lose much of the information related to the input, hash functions for example.
Yes you could use a hash function as round function, but if you are using the "same key" over all rounds, you are vulnerable to slide attacks. Using a hash function is not a very good idea. 

Your round function should not introduce biases, should not lead to special differences (attack: differential cryptanalysis), and it should also not be writable as a linear equation (attack: linear cryptanalysis). You should take care of how your round function will work when the round keys are slightly modified (attack: related key attacks), and you should take care of impossible differentials. If you got all that right, you can use any $F$ which passed these criterions.

Short answer is, $F$ needs to be carefully selected. And “yes” you are interested, that $F$ will "loose as much information about the input" as possible.

Your second question:


  This question might seem silly, but how true is the fact that the round function $F$ does not have to be invertible?


Your round function does not need to be invertible, because the classic balanced Feistel construction will create two halves, where only one half is modified, and the other kept in clear text/unmodified. After each round both halfes are exchanged and the modified half becomes the fixed one and the previously unmodified half is now encrypted. If you now reverse the order of the keys/rounds you can decrypt the message again. The Feistel construction itelf is invertible.