Select sub-sequences in lists?

I was expecting mathematica to also be able to access the fields in place without writing a new list

In-place modification approach:

list = {{a1, {b1, c1}}, {a2, {b2, c2}}, {a3, {b3, c3}}};
list[[;; , 2]] = list[[;; , 2, 2]];
list

{{a1, c1}, {a2, c2}, {a3, c3}}

Other solutions

MapAt[Last, list, {All, 2}]
(* {{a1, c1}, {a2, c2}, {a3, c3}} *)

list /. {x_, {y_, z_}} -> {x, z}
(* {{a1, c1}, {a2, c2}, {a3, c3}} *)

Benchmark

Large data, 10^7 lines.

largeList = 
  Table[{RandomReal[1], {RandomReal[1], RandomReal[1]}}, {10^7}];

First@AbsoluteTiming returns the absolute number of seconds in real time that have elapsed during evaluation.

First@AbsoluteTiming[
  MapAt[Last, largeList, {All, 2}]
  ]
(* 4.86435 *)

First@AbsoluteTiming[
  largeList /. {x_, {y_, z_}} -> {x, z}
  ]
(* 4.95776 *)

First@AbsoluteTiming[
  Partition[Flatten@largeList, 3][[All, {1, 3}]]
  ]
(* 3.66688 *)

First@AbsoluteTiming[
  {#, #2[[-1]]} & @@@ largeList
  ]
(* 8.83799 *)

First@AbsoluteTiming[
  {#[[1]], #[[2, -1]]} & /@ largeList
  ]
(* 12.3966 *)

First@AbsoluteTiming[
  Module[{modlist = largeList}
   , modlist[[;; , 2]] = modlist[[;; , 2, 2]]
   ]
  ]
(* 2.6865 *)

First@AbsoluteTiming[
  Query[All, {1 -> Identity, 2 -> Last}]@largeList
  ]
(* 23.1379 *)

First@AbsoluteTiming[
  {#, #2 & @@ #2} & @@@ largeList
  ]
(* 12.7286 *)

First@AbsoluteTiming[
  {#[[1]], #[[2, 2]]} & /@ largeList
  ]
(* 13.2996 *)

First@AbsoluteTiming[
  largeList[[;; , 2]] = largeList[[;; , 2, 2]]
  ]
(* 2.07737 *)

Update: Using a slightly modified version of the list of parts:

list = {{a1, {b1, c1}}, {a2, {b2, c2}}, {a3, {b3, c3}}};
parts = {{1}, {2, 2}};
list2 = {{d1, e1}, {d2, e2}, {d3, e3}};
Do[(list[[i, ##]] = list2[[i, #]]) & @@@ parts, {i, Length @ list2}]
list

{{d1, {b1, e1}}, {d2, {b2, e2}}, {d3, {b3, e3}}}

Alternatively, you can use ReplacePart:

list = {{a1, {b1, c1}}, {a2, {b2, c2}}, {a3, {b3, c3}}};
list = ReplacePart[list, ({i_, ##} :> list2[[i, #]]) & @@@ parts];
list

{{d1, {b1, e1}}, {d2, {b2, e2}}, {d3, {b3, e3}}}

Original answer:

Few additional alternatives:

Transpose@{list[[All, 1]], list[[All, 2, 2]]}
Cases[list , {a_, {b_, c_}} :> {a, c}]
Replace[list , {_, b_} :> b, {2}]
Query[All, {1 -> Identity, 2 -> Last}] @ list
{#, #2 & @@ #2} & @@@ list
{#, #2[[-1]]} & @@@ list
{#[[1]], #[[2, 2]]} & /@ list 

all give

{{a1, c1}, {a2, c2}, {a3, c3}}