Vtyjkyuk

Define function f

Clear[f, x];
f[x_] := 2^x;

N[ f[arg], #sig. digits]

data = Table[
 N[Pi, n],
 N[ f[N[Pi, n]], 10 ]
 , n, 1, 8
 ];

Text gridding...

Text@Grid[Prepend[data, "x", "f(x)"],
 Alignment -> Left,
 Dividers -> Center, 2 -> True
 ]

beginarrayl
textx & textf(x) \
hline
3. & 0. \
3.1 & 9. \
3.14 & 8.8 \
3.142 & 8.82 \
3.1416 & 8.825 \
3.14159 & 8.8250 \
3.141593 & 8.82498 \
3.1415927 & 8.824978 \
endarray

How do I show the following instead?

beginarrayl
textx & textf(x) \
hline
3. & 8.000000 \
3.1 & 8.574188 \
3.14 & 8.815241 \
3.142 & 8.821353 \
3.1416 & 8.824411 \
3.14159 & 8.824962 \
3.141593 & 8.824974 \
3.1415927 & 8.824978 \
endarray

Using arbitrary precision arithmetic produces the 1st result you show because Mathematica normally does not show digits with no precision since they are just noise. However, if you are trying to show how having better and better rational approximations to π improves the approximation of $2^pi$, I suggest the following approach.

data =
 With[x = π, n = 8,
 Table[
 With[u = Round[x, 10^-i], u, Round[f[u], 10^-i]] // N,
 i, 0, n]];
TableForm[Map[NumberForm[#, 10, 8] &, data, 2],
 TableHeadings -> None, x, f[x]]

It might take some additional formatting to get exactly what you're after, but this is a start. I acknowledge some things may be able to be made simpler.

Clear[f, x];
f[x_] := 2^x;
data = N@Table[
 Table[10^(-j + 1), j, 1, i].RealDigits[N@Pi, 10, i][[1]], i, 
 8] /. x_?NumericQ :> NumberForm[x, 8], f[x];

Text@Grid[Prepend[data, "x", "f(x)"], Alignment -> Left, 
 Dividers -> Center, 2 -> True]