JavaScript (ES6), 36 bytes

f=([c,...r],s=`

`)=>c?s+f(r,s+c)+s:s

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Python 2, 60 52 bytes

f=lambda s,n=1:s[n:]and[s[:n]]+f(s,n+1)+[s[:n]]or[s]

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Python 3.8 (pre-release), 50 bytes

f=lambda s,n=1:s>(x:=s[:n])and[x,*f(s,n+1),x]or[s]

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Perl 6, 31 bytes

{[~](@_)[0...@_-1...0]}o*.comb

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Anonymous code block that takes a string and returns a list of lines.

05AB1E (legacy),  4  3 bytes

^{Crossed out  4  is no longer 4 :)}

η.∊

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Explanation:

η     # Get the prefixes of the (implicit) input-string

 .∊   # Vertically mirror everything with the last line overlapping

      # (which implicitly joins by newlines in the legacy version of 05AB1E)

      # (and output the result implicitly)

In the new version of 05AB1E, and explicit » is required after the η, which is why I use the legacy version of 05AB1E here to save a byte.

Japt, 9 7 bytes

-2 bytes thanks to Shaggy

Êõ@¯XÃê

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C# (Visual C# Interactive Compiler), 123 109 94 84 74 bytes

Assumes we can return a char array array (I believe we can, as a char array is a valid representation for a string and a string array is a valid representation for multiple lines)

a=>new int[a.Length*2-1].Select((b,i)=>a.SkipLast(Math.Abs(a.Length-i-1)))

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JavaScript (Node.js), 90 bytes

This can probably be golfed alot more, Arnauld already has a way shorter one but I had fun atleast!

s=>{a=[];for(c=s.length-1;c--;)a[c]=s.slice(0,c+1);return[...a,s,...a.reverse()].join`n`}

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Japt -R, 4 bytes

å+ ê

Cumulative reduce on a string.

-1 byte thanks to @Shaggy

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C (gcc), 68 67 bytes

i,j,t;f(char*s){for(;j?--i:++i;puts(s),j+=!t,s[i]=t)t=s[i],s[i]=0;}

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MATL, 8 bytes

nZv"G@:)

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Please like this post for the smiley :) in the code it took me a lot of time to make.

n  % Length of the input string

Zv % Symmetric range ([1 2 ... n ... 1])

"  % For each k in above range

G  % Push input

@: % Push [1 2 ... k]

)  % Index

APL (Dyalog Unicode), 9 bytes^SBCS

Anonymous tacit prefix function. Returns list of strings.

(⊢,1↓⌽),

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, the list of prefixes (lit, the cumulative concatenation)

(…) apply the following function to that list:

 ⌽ the reversed list

 1↓ drop the first item

 , prepend

 ⊢ the unmodified list

Charcoal, 5 bytes

Ｇ^Ｌθθ

Try it online! Link is to verbose version of code. Explanation: Ｇ draws a filled polygon, ^ specifies that the sides are down right and down left (the polygon then automatically closes itself), Ｌθ specifies the length of those sides as being the length of the original input and the final θ specifies the fill string.

Perl 5 (-p), 26 bytes

s,.,$=$`.$/.$;"$`$&

",ge

TIO

Ruby, 51 42 40 bytes

f=->s,i=1{s[i]?[t=s[0,i],*f[s,i+1],t]:s}

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Thanks to Doorknob for -2 bytes.

J, 11 bytes

Anonymous tacit prefix function. Returns a space-padded character matrix.

[:(}:,|.)]

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] the list of prefixes

[:(…) apply the following function to that list

 |. the reverse list

 , prepended with

 }: the curtailed (without last item) list

C# (Visual C# Interactive Compiler), 82 bytes

x=>{int l=x.Length,i=0;for(;i<l*2-1;)WriteLine(x.Substring(0,l-Math.Abs(++i-l)));}

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String as input and output to std

J, 12 bytes

],[:}.@|.]

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Still 1 byte longer than Adám's

Initial solution:

J, 17 16 bytes

{."{~#-|@i:@<:@#

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An example REPL session:

   # 'String'

6

   i:#'String'                            

_6 _5 _4 _3 _2 _1 0 1 2 3 4 5 6

   i:<:#'String'                          

_5 _4 _3 _2 _1 0 1 2 3 4 5

   |i:<:#'String'                         

5 4 3 2 1 0 1 2 3 4 5

   (#'String')-|i:<:#'String'             

1 2 3 4 5 6 5 4 3 2 1

   'String'{."0 1~(#'String')-|i:<:#'String' 

S     

St    

Str   

Stri  

Strin 

String

Strin 

Stri  

Str   

St    

S     

Julia, 44 bytes

s->(a=cumprod([s...]);[a;reverse(a)[2:end]])

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Gets the first half by taking the cumulative product (* is string concatenation in Julia) of the array of characters, then adds this array to itself reversed minus the first element.

Octave, 58 bytes

for k=1:(n=nnz(s=input(''))*2)-1

disp(s(1:min(k,n-k)))

end

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Red, 78 bytes

func[s][repeat n l: length? s[print copy/part s n]loop l[take/last s print s]]

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Brachylog, 8 bytes

a₀ᶠ;Lc.↔

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Explanation

a₀ᶠ          Find all prefixes of the input

   ;Lc.      The output is that list of prefixes with something unknown appended at the end

      .↔     The output reversed is itself (i.e. it's a palindrome)

R, 86 bytes

x=utf8ToInt(scan(,''))

for(i in c(y<-1:sum(x|1),rev(y)[-1]))cat(intToUtf8(x[1:i]),"

")

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I'm learning more and more about better ways to manipulate strings in R, so I'm somewhat proud of this answer. The only part I don't like is the for loop portion, which I feel could definitely be golfed.

MBASIC, 103 bytes

1 INPUT S$:N=1

2 PRINT LEFT$(S$,N):IF N<LEN(S$) THEN N=N+1:GOTO 2

3 N=N-1:PRINT LEFT$(S$,N):IF N>1 THEN 3

PowerShell, 89 bytes

param($a)0..($d=$a.Length-1)|%{[string[]]$b+=,$a[0..$_]};($c=$b-replace' ')+$c[($d-1)..0]

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Pyke, 6 bytes

BE 27 4F 5F 2B 58

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        - input() (implicit)

.>      - prefixes(^) (1 byte)

  'O_   - ^[:-1], reversed(^)

     +  - [^]

      X - "n".join(reversed(^))