What If the $g_{rr}$ goes to infinity but the $g_{tt}$ stays finite at some radius or vice versa? I am not that familiarized with general relativity so this question might sound obvious, but I was wondering what exactly the metric components on a curved spacetime represent. For instance what does it mean when  $g_{rr}\rightarrow\infty$ while $g_{tt}$ remains finite? Or what if the inverse thing occurs, and $g_{tt}\rightarrow\infty$ while $g_{rr}$ is finite? In the second case would the solution represent a black hole or not?
You could consider the static, spherically symmetric spacetime:
$$ds^2=-e^{A(r)}dt^2+e^{B(r)}dt^2+r^2(d\theta^2+\sin^2\theta \,d\phi^2)$$ 
By adjusting your generalized gravity theory it is possible to come up with a case where after solving the equations of motion, the functions $A(r),\,B(r)$ are found so that at some distance only one of them goes to infinity.
 A: If any of the components of $g_{ab}$ or $g^{ab}$ goes infinite, then you inevitably must have either a coordinate or curvature singularity at that point.  If you can find a coordinate transformation that removes the infinity, then you merely have a coordinate singularity.
The presence of infinities in the metric coordinates is independent of the presence of a black hole.  there are many definitions of black holes, but almost all of them are based on whether light can or can't escape a closed surface.  Meanwhile, introducing coordinate singularites is easy.
Take the 2-D Minkowski metric:
$$ds^{2} = -dt^2 + dx^{2}$$
now, do the transformation:
$$t = \frac{1}{T^{3}}\;\;\;x=\frac{1}{1-X}$$
Then:
$$ds^{2} = -\frac{9dT^2}{T^8} + \frac{dX^{2}}{\left(1-X\right)^{2}}$$
which has a singularity at $T=0$ and $X=1$, but these are just functions of our coordinate choice
A: The unique spherically symmetric vacuum solution in general relativity (GR) is the Schwarzchild solution, with 
$\exp{A}$ = 1 - $r_s$/r
where $r_s$ is the Schwarzschild radius, the horizon for a black hole (BH), and B = -A
Indeed the first term (temporal from the outside coming in) is 0 at the horizon, and the second (radial term) is infinite. It's not a singularity just the event horizon. 
See the solution metric at https://en.m.wikipedia.org/wiki/Schwarzschild_metric
See also the extended solutions showing that the Schwarzschild solution can be transformed into other coordinates that makes it clear that the horizon is just a coordinate problem (coordinate singularity) using the Schwarzschild metric, but it is not a real singularity of spacetime because you can find other coordinates that don't blow up at the horizon. An example (and there are others, also used) is the Kruskal-Szekeres coordinates. See it at https://en.m.wikipedia.org/wiki/Kruskal–Szekeres_coordinates. That and some of the others do not blow up anywhere except at the real spacetime (curvature) singularity at r=0. They are useful to explore and understand the full possible geometry of spacetime for BHs. You can see also the White Holes on the other side. Whether those are physically real or not remains unknown. 
So you don't need to modify gravity from GR to find the example you wanted. Even if not in the forms you wanted (exponentials) you can clearly make the transformations to an exponential form for the first two metric components from the Schwarzschild metric (having the exponential of the log of the terms). 
So, those at least, are indeed BHs, with the first term (temporal outside, radial inside) zero at the horizon and infinite at the real singularity, and the second term (radial on the outside, the opposite inside) infinite at the horizon and zero at the singularity. At the horizon time dilates to infinity for a still observer at infinity, so he/she never sees anything falling into the horizon. At the singularity at r=0 the covariant curvature becomes infinite. 
When you do have term in the metric which becomes infinite it may or not be a real singularity. You have to calculate the invariant curvature from invariants using the Riemman tensor to find out, or find a coordinate systems which does not blow up there, or calculate geodesics and see if they suddenly end at some point/surface/hypersurface. It is not often obvious. 
Now, the situation is different for non-vacuum solutions. There are static spherically symmetric solutions for perfect fluid models. See for instance the family of possibilities for the Tolman-Oppenheimer-Volkoff models at https://en.m.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation. They don't need to be BHs, but will collapse for certain limits. 
There are also indeed static spherically symmetric solutions in alternative theories of gravity. See for instance for f(R) gravity http://iopscience.iop.org/article/10.1088/0264-9381/24/8/013/meta
