Why is it common practice to create a hash of the message and sign that instead of signing the message directly?
Well, the RSA operation can't handle messages longer than the modulus size. That means that if you have a 2048 bit RSA key, you would be unable to directly sign any messages longer than 256 bytes long (and even that would have problems, because of lack of padding).
In contrast, a cryptographical hash can take an arbitrarily long message, and 'compress' it into a short string, in such a way that we cannot find two messages that hash to the same value. Hence, signing the hash is just as good as signing the original message; without the length restrictions we would have if we didn't use a hash.
The important part and this is where I really started scratching my head: How can the recipient verify that I own the private key if the public key seems to be enough to recreate the signature?
What made you think that the public key is enough to recreate the signature? It is sufficient to verify a signature that you're given, but it is not sufficient to generate new ones (or so we hope; if that's not true, the signature scheme is broken).
If you're using RSA, the signature verification process is (effectively) checking whether:
$S^e = Pad( Hash( M ))\ \ (\bmod \ N)$
(where $S$ is the signature, $M$ is the message, and $e$ and $N$ are parameters from the public key; I say "effectively" because sometimes the padding method is nondetermanistic; that makes this check slightly different, but not in a way that matters for this discussion).
Now, if we were trying to forge a signature for a message $M'$ (with only the public key), we could certainly compute $P' = Pad(Hash(M'))$; however, then we'd need to find a value $S'$ with:
$S'^e = P' \ (\bmod\ N)$
and, if $N$ is an RSA modulus, we don't know how to do that.
The holder of the private key can do this, because he has a value $d$ with the property that:
$(x^e)^d = x \ (\bmod\ N)$
for all $x$. That means that:
$(P')^d = (S'^e)^d = S' \ (\bmod\ N)$
is the signature.
Now, if we have only the public key, we don't know $d$; getting that value is equivalent to factoring $N$, and we can't do that. The holder of the private key knows $d$, because he knows the factorization of $N$.