Why can't the Higgs be produced with a single gluon? In the PDG there are 4 important production mechanisms listed (page 11). Why is it that for the first and fourth mechanism a second gluon is needed? 
For the first one, why can't the gluon decay into $t\bar t$ which then continue to annihilate into a $H$? 
For the fourth, why can't the gluon decay into $t \bar t$ where one $t$-Quark radiates off a $H$?
 A: There are rules for Feynman diagrams, dictated from energy and momentum conservation and conservation of various quantum numbers, and also the coupling constants allowed.

Why can't the Higgs be produced with a single gluon?

You need  loops of quarks or leptons because gluons do not couple with the weak coupling constant, and cannot couple directly to the Higgs; 
The diagrams in the pdg are not complete, one has to supply the energy for the gluons and quarks in a definite real particle input. Both quarks and gluons are off mass shell in these diagrams, which are intermediate to the measurement of a real Higgs particle.
Here are the diagrams :

Let us see your questions:

For the first one, why can't the gluon decay into $t \bar t$ which then continue to annihilate into a H ? 

If you extend on the left diagram c with an  incoming one gluon in a loop with $q \bar q$ you let q be t. It will be very suppressed as a diagram due to the mass of the t.

For the fourth, why can't the gluon decay into $t \bar t$ where one t-Quark radiates off a H ?

The outgoing are on mass shell , with a lot of energy going to the masses. For one of the top to radiate an H the probability will be very small because of the masses in the  necessary  off mass shell loops , otherwise it will be just top decay to Higgs+X which is a very rare process in the standard model anyway. 
A: The first reaction you propose,
$$g \to t\bar{t} \text{ loop} \to H$$
violates the conservation of color charge, because a gluon has color and a Higgs doesn't. It also violates the conservation of angular momentum, because a gluon has spin 1 and a Higgs has spin 0. Finally if the gluon is on-shell (which it isn't in this context) it would also violate the conservation of energy-momentum.
The first two points may be a bit confusing because the processes $g \to t \bar{t}$ and $t \bar{t} \to H$ are both allowed. But there are constraints here: the tops in the first example have to have net color and net spin, while the tops in the second example can't.
I'm not totally sure about the second reaction you propose,
$$g \to t \bar{t} H$$
but I'm going to post an answer anyway, because the best way to learn the right answer is to post something wrong. I suspect the issue is that the initial gluon must be very far off-shell, which means the reaction is extremely rare by the asymptotic freedom of QCD. (When you have two initial gluons, you don't need them off-shell at all; two on-shell gluons coming in back-to-back have plenty of invariant mass.) This reasoning applies to the first process as well.
