The simplest way to explain the Christoffel symbol is to look at them in flat space. Normally, the laplacian of a scalar in three flat dimensions is:
$$\nabla^{a}\nabla_{a}\phi = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$$
But, that isn't the case if I switch from the $(x,y,z)$ coordinate system to cylindrical coordinates $(r,\theta,z)$. Now, the laplacian becomes:
$$\nabla^{a}\nabla_{a}\phi=\frac{\partial^{2}\phi}{\partial r^{2}}+\frac{1}{r^{2}}\left(\frac{\partial^{2}\phi}{\partial \theta^{2}}\right)+\frac{\partial^{2}\phi}{\partial z^{2}}-\frac{1}{r}\left(\frac{\partial\phi}{\partial r}\right)$$
The most important thing to note is the last term above--you now have not only second derivatives of $\phi$, but you also now have a term involving a first derivative of $\phi$. This is precisely what a Christoffel symbol does. In general, the Laplacian operator is:
$$\nabla_{a}\nabla^{a}\phi = g^{ab}\partial_{a}\partial_{b}\phi - g^{ab}\Gamma_{ab}{}^{c}\partial_{c}\phi$$
In the case of cylindrical coordinates, what the extra term does is encode the fact that the coordinate system isn't homogenous into the derivative operator--surfaces at constant $r$ are much larger far from the origin than they are close to the origin. In the case of a curved space(time), what the Christoffel symbols do is explain the inhomogenities/curvature/whatever of the space(time) itself.
As far as the curvature tensors--they are contractions of each other. The Riemann tensor is simply an anticommutator of derivative operators--$R_{abc}{}^{d}\omega_{d} \equiv \nabla_{a}\nabla_{b}\omega_{c} - \nabla_{b}\nabla_{a} \omega_{c}$. It measures how parallel translation of a vector/one-form differs if you go in direction 1 and then direction 2 or in the opposite order. The Riemann tensor is an unwieldy thing to work with, however, having four indices. It turns out that it is antisymmetric on the first two and last two indices, however, so there is in fact only a single contraction (contraction=multiply by the metric tensor and sum over all indices) one can make on it, $g^{ab}R_{acbd}=R_{cd}$, and this defines the Ricci tensor. The Ricci scalar is just a further contraction of this, $R=g^{ab}R_{ab}$.
Now, due to Special Relativity, Einstein already knew that matter had to be represented by a two-index tensor that combined the pressures, currents, and densities of the matter distribution. This matter distribution, if physically meaningful, should also satisfy a continuity equation: $\nabla_{a}T^{ab}=0$, which basically says that matter is neither created nor destroyed in the distribution, and that the time rate of change in a current is the gradient of pressure. When Einstein was writing his field equations down, he wanted some quantity created from the metric tensor that also satisfied this (call it $G^{ab}$) to set equal to $T^{ab}$. But this means that $\nabla_{a}G^{ab} =0$. It turns out that there is only one such combination of terms involving first and second derivatives of the metric tensor: $R_{ab} - \frac{1}{2}Rg_{ab} + \Lambda g_{ab}$, where $\Lambda$ is an arbitrary constant. So, this is what Einstein picked for his field equation.
Now, $R_{ab}$ has the same number of indicies as the stress-energy tensor. So, a hand-wavey way of looking at what $R_{ab}$ means is to say that it tells you the "part of the curvature" that derives from the presence of matter. Where does this leave the remaining components of $R_{abc}{}^{d}$ on which $R_{ab}$ does not depend? Well, the simplest way (not COMPLETELY correct, but simplest) is to call these the parts of the curvature derived from the dynamics of the gravitational field itself--an empty spacetime containing only gravitational radiation, for example, will satisfy $R_{ab}=0$ but will also have $R_{abc}{}^{d}\neq 0$. Same for a spacetime containing only a black hole. These extra components of $R_{abc}{}^{d}$ give you the information about the gravitational dynamics of the spacetime, independent of what matter the spacetime contains.
This is getting long, so I'll leave this at that.
UPDATE
This answer raised some controversy, which I believe comes from the fact I didn't properly define what I mean by energy. In this post, whenever I say energy I refer to the canonical/Hilbert definition of the energy-momentum tensor, i.e., the functional derivative of the Lagrangian with respect to the metric. In GR, this is the energy that matters, because it directly enters in the EFE.
Note that this definition, in principle, has nothing to do with the classical interpretation of energy(=work), which is conserved along the trajectories of point particles. In GR we don't care about this latter energy, because it is not conserved in general (you need some Killing fields to exist, but OP asked us not to mention Killing fields).
Finally, in some semi-classical analysis of GR we can define some kind of potential energy (e.g., $2\phi(r)\sim 1+g_{00}(r)$), which does behaves like the classical interpretation of energy(=work). But this semi-cassical analysis, where you mix Newtonian and GR concepts, can only be used for weak or highly symmetric fields.
With this in mind, note that you can have bodies accelerating/slowing down because of the action of gravity. This doesn't contradict the fact that gravitational energy doesn't exist: when we say energy is conserved in GR, we mean $\nabla T=0$, but this energy is the canonical/Hilbert energy, not the kinetic+potential energy of the bodies! If you want to talk about kinetic+potential energy of a test body, you'll need Killing sooner or later!
First Question
Why should the EMT even contain gravitational energy? Wouldn't this energy be proportional to curvature? This energy would than again cause additional curvature which would in turn increase the grav. energy which would again increase the curvature etc.
The Einstein Field Equations tell you how curvature is related to energy:
$$
R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=T_{\mu\nu}
$$
(where I use natural units $c=8\pi G=1$)
The l.h.s. is a measure of curvature, while the r.h.s. is a measure of the energy of the system. Once you know the r.h.s, you can solve these PDEs to find the metric (l.h.s).
Some examples:
You probably know from classical electromagnetism that the electromagnetic field carries energy (something like $\boldsymbol E^2+\boldsymbol B^2$ should look familiar). In tensor notation, the energy is written
$$
T^{\mu\nu}=F^{\mu\alpha}F^\nu_\alpha-\frac{1}{4}g^{\mu\nu}F^2
$$
where $F$ is the electromagnetic strenth tensor (note that $T^{00}\propto \boldsymbol E^2+\boldsymbol B^2$).
If you have some matter around that you can model as a perfect fluid (e.g, negligible viscosity), then the formula for the energy is given by
$$
T^{\mu\nu}=(\rho+p)u^\mu u^\nu+p g^{\mu\nu}
$$
where $\rho,p$ are the mass density and the pressure of the fluid, and $u$ its velocity.
Etc
Anything that is present in the system generates gravity, and thus you should include its energy in the EFE: $T^{\mu\nu}=T_1^{\mu\nu}+T_2^{\mu\nu}+\cdots$, where each $T_i^{\mu\nu}$ is a different source of energy. But we dont include a term $\boldsymbol{T_i^{\mu\nu}}$ for curvature: gravity has no energy tensor:
$$
\text{r.h.s}=T_\text{EM}+T_\text{matter}+T_\text{quantum?}+\cdots+ \overbrace{ T_\text{gravity}}^{\text{NO!}}
$$
There is no tensor for gravitational energy and there is no term for gravity in the r.h.s. of the the EFE. In the r.h.s. of the EFE you only include nongravitational forms of energy.
Note that when in vacuum (i.e. no matter/no radiation), the EFE are
$$
R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=0
$$
You can clearly see that there is no term in the r.h.s., even though there is gravity.
Therefore, the answer to your first question is: the EMT should not contain gravitational energy.
Second Question
What is gravitational energy in GR? (a formula and/or picture would be nice)
In GR it doesn't make sense to speak of gravitational energy.
Short explanation: as you probably know, in GR gravity is not a true force$^1$. Therefore, there is no potential energy$^2$ associated to it.
Longer explanation: in GR energy is defined as
$$
T^{\mu\nu}=\frac{-2}{\sqrt{|g|}}\frac{\delta \mathcal L_m}{\delta g^{\mu\nu}}
$$
where
$$
\mathcal L=\sqrt{|g|}\left(R+\mathcal L_m\right)
$$
is the (full) Lagrangian of the theory, and $\mathcal L_m$ is the nongravitational part of the Lagrangian. As you can see, the energy is defined through the nongravitational part of the Lagrangian, which in turns means that it simply makes no sense to speak of the gravitational energy.
Note that it is possible to define a quantity that (supposedly) looks a lot like a gravitational energy$^3$:
$$
t^{\mu\nu}=\frac{1}{2}Rg_{\mu\nu}-R_{\mu\nu}+\frac{1}{2g}[g(g^{\mu\nu}g^{\alpha\beta}-g^{\mu\alpha}g^{\nu\beta})]_{,\alpha\beta}
$$
but it is not in general accepted as a true gravitational energy (it's more like a formal analogy, without much use AFAIK). Its not even a tensor. However, to see how it works, we can calculate $t^{00}$ in the Schwarzschild solution in the weak field (non-relativistic) limit:
$$
g^{00}=-\left(1-\frac{2M}{r}\right) \qquad g^{ij}=\delta^{ij}
$$
(where now I take $c=G=1$)
The "potential energy" of this metric is
$$
t^{00}=\frac{1}{2g} \nabla^2 (g^2)=\nabla^2g+\frac{1}{2g}(\nabla g)^2
$$
where $g=g^{00}$. I wonder if this looks like a potential energy to you (IMHO, it does not).
EDIT: There is a certain point I'd like to discuss a bit further. It just happens that in EFE we don't include curvature-energy in the r.h.s. You point out that if this were the case, then the curvature would cause more curvature, and the new curvature would cause even more, etc. Is this possible? or is it just absurd?
Well, I think you will like a lot this lecture by Feynman (Vol II. Chap 23: Cavity Resonators). The r.h.s. of Maxwell equations include both $\boldsymbol E$ and $\boldsymbol B$, so in this case there exists the feedback you talk about: a certain electric field can be responsible for a magnetic field, which in turns generates more electric field, which is responsible for more magnetic field, ad infinitum. But the result is a convergent series, so everything works out just fine (I know this is kinda irrelevant for your question, but I think its neat, and I believe you might like it as well :) )
$^1$: "In general relativity, the effects of gravitation are ascribed to spacetime curvature instead of a force." - from Wikipedia.
$^2$: "In physics, potential energy is the energy that an object has due to its position in a force field [...]" - from Wikipedia.
$^3$: this quantity satisfies certain relations that energy usually satisfies, and it's constructed using just the metric tensor $g$.
Best Answer
The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).
Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the ansatz already imposes Lorentz signature.