Sorry im really not great with mathematica, and it's along the right lines, basically my result is the following

But i was hoping i could have things such as the the 2nd to last line, the 3rd term, ideally id like it to read (sin2?)e?2Y(?1 +e2Y?r?rX+r?rY) instead, i know its correct but for my future calculations it would be nice if in my appendix, the code would print exactly that, thankyou for your time

Well, of course you have to execute the whole code - I posted above the changed part only! OK, it then reads:

ClearAll["Global`*"]
d = 4; y = {t, r, \[Theta], \[Phi]};
G = {{-(E^(2 X[t, r])), 0, 0, 0}, {0, (E^(2 Y[t, r])), 0, 0}, {0, 0, 
    r^2, 0}, {0, 0, 0, r^2 Sin[\[Theta]]^2}};
IG = Inverse[G];
\[CapitalGamma] = 
  Inactivate[
   Table[1/2*
     Sum[IG[[i, l]]*(D[G[[j, l]], y[[k]]] + D[G[[k, l]], y[[j]]] - 
         D[G[[j, k]], y[[l]]]), {l, 1, d}], {i, 1, d}, {j, 1, d}, {k, 
     1, d}], D];
Rct = Inactivate[
   Table[D[\[CapitalGamma][[i, l, j]], y[[k]]] - 
     D[\[CapitalGamma][[i, l, k]], y[[j]]] + 
     Sum[\[CapitalGamma][[i, k, a]]*\[CapitalGamma][[a, l, j]], {a, 1,
        d}] - Sum[\[CapitalGamma][[i, l, b]]*\[CapitalGamma][[b, k, 
         j]], {b, 1, d}], {i, 1, d}, {j, 1, d}, {k, 1, d}, {l, 1, d}],
    D];
rct = Table[Sum[Rct[[c, i, c, j]], {c, 1, d}], {i, 1, d}, {j, 1, d}];
rct

This gives a lengthy result, but with unresolved partial derivatives. Is it this you want?

Maybe Inactivate[<some expression>, D] is what you are looking for:

\[CapitalGamma] = 
  Inactivate[
   Table[1/2*
     Sum[IG[[i, l]]*(D[G[[j, l]], y[[k]]] + D[G[[k, l]], y[[j]]] - 
         D[G[[j, k]], y[[l]]]), {l, 1, d}], {i, 1, d}, {j, 1, d}, {k, 
     1, d}], D];
Rct = Inactivate[
   Table[D[\[CapitalGamma][[i, l, j]], y[[k]]] - 
     D[\[CapitalGamma][[i, l, k]], y[[j]]] + 
     Sum[\[CapitalGamma][[i, k, a]]*\[CapitalGamma][[a, l, j]], {a, 1,
        d}] - Sum[\[CapitalGamma][[i, l, b]]*\[CapitalGamma][[b, k, 
         j]], {b, 1, d}], {i, 1, d}, {j, 1, d}, {k, 1, d}, {l, 1, d}],
    D];