Perl 6, 25 bytes

->a,b{b~~/(10|.)**{a}/}

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Anonymous code block that takes a number and a string and returns as a Match object.

Explanation:

->a,b{                }  # Anonymous code block taking params a and b

        b~~/           /   # Match using b

            (10|.)           # 10 or a single digit

                  **{a}      # Exactly a times, being greedy

Python 3, 47 bytes

lambda s,n:[*s.replace(b'1',b'n',len(s)-n)]

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Takes the "one line" as a bytestring with raw bytes x00 - x09. If it's not acceptable:

Python 3, 56 bytes

lambda s,n:[x-48for x in s.replace(b'10',b':',len(s)-n)]

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Takes "one line" as bytestring.

Brachylog, 23 21 bytes

-2 bytes thanks to Fatalize

h~c.{ịℕ≤10&ịṫ?∧}ᵛ&t~l

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The input is a pair [Line, N].

This is my first Brachylog program, so there is probably a lot room for improvement.

It is very slow when the length of the line > 7.

Explanation:

h~c.{ịℕ≤10&ịṫ?∧}ᵛ&t~l

h                         The first element in the input

 ~c                       is formed by concatenating

   .                      the elements in the output array

   .{         ∧}ᵛ     AND For every element in the output array holds that

     ị                      The element converted to an integer

      ℕ                       is a natural number

       ≤10                    and less than or equal to 10

          &ịṫ?              and it has no leading zeroes (*)

                 &t   AND The second element of the input

                   ~l     is the length of the output 

(*) ịṫ? checks that there are no leading zeroes. It converts the string to integer and then back to string and compares to the original string.

V, 17, 12 bytes

ÓòÀGjí1“î…0

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I was content with 17 bytes, but than 05AB1E came along with 13, and I couldn't let a challenge go unanswered. :D

Ó                      " Put each character on it's own line

  ò                     " Recursively (repeat until an error happens)...

   ÀG                   "   Go to the "n"th line

     j                  "   Move down a line (this will error if there are exactly "n" lines)

      í                 "   Remove...

       1                "     a '1'

        <0x93>          "     START THE MATCH HERE

              î         "     a newline

               <0x85>   "     END THE MATCH HERE

                   0    "     a '0'

Hexdump:

00000000: 5cd3 f2c0 476a ed31 93ee 8530            ...Gj.1...0

Alternate solution:

ÓòÀGjç1î0/J

Unfortunately, this replaces 10 with 1 0

Ruby, 57 bytes

->n,m{m.sub!"10",?A while m[n];m.chars.map{|c|c.to_i 16}}

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This may turn out to be not the golfiest approach, but it looks like a fun idea to temporarily substitute 10 for a hex A, which incidentally is also a high mark (if we consider A-F grading system :))

JavaScript, 57 52 bytes

n=>g=s=>s[n]?g(s.replace(x=10,`x`)):[...s].map(eval)

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Python 3, 71 68 59 bytes

down another 9 bytes thanks to ovs.

lambda n,s:[int(c,11)for c in s.replace('10','a',len(s)-n)]

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I was iniitially trying to use str.partition() recursively, but using replace smacked me in the face not too long after. Can anyone improve on this?

Also, here's a TIO link that I used to make the test cases into something more copy/pasteable

Haskell, 98 bytes

n!x=[y|y<-s x,y==take n y]!!0

s('1':'0':x)=do y<-s x;[1:0:y,10:y]

s(x:y)=(read[x]:)<$>s y

s _=[]

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Explanation

The function s does all possible splits, for example: "1010" becomes [[1,0,1,0],[10,1,0],[1,0,10],[10,10]], note how the longest splits end up at the beginning (because 1:0:y comes before 10:y).

With that in mind, we can take all these values and filter the ys out where y == take n y which keeps also splits that are shorter than required. For example with 4 we leave the list the same [[1,0,1,0],[10,1,0],[1,0,10],[10,10]].

Now we can just get the first element in that list because the inputs will always be valid (eg. 5!"1010" would give [1,0,1,0] too, but we don't need to handle it).

Note: I somehow miscounted.. y==take n y is the same length as length y==n :S

Haskell, 68 bytes

n!('1':'0':x)|n-2<length x=10:(n-1)!x

n!(s:x)=read[s]:(n-1)!x

n!_=

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Greedily take 10s as long as there are more digits than marks remaining.

Perl 5 -plF, 39 bytes

$a=<>;$_="@F";s/1 0/10/ while$a-1<y/ //

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Clean, 128 bytes

import StdEnv

@=[]

@['10':t]=[u++v\u<-[[10],[1,0]],v<- @t];@[h:t]=[[digitToInt h:v]\v<- @t]

?n l=hd[e\e<- @l|length e==n]

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05AB1E, 13 bytes

.œsù.ΔïTÝÃJ¹Q

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or as a Test Suite

Explanation

.œ              # partitions of the first input

  sù            # of a length equal to the second input

    .Δ          # find the first partition that returns true when:

      ï         # each element is converted to integer

       TÝÃ      # and only numbers in [0 ... 10] are kept

          J     # then join it together

           ¹Q   # and compare it to the first input for equality

JavaScript (Babel Node),  70 69  59 bytes

Takes input as (n)(line).

n=>s=>(a=s.match(/10|./g)).flatMap(x=>x>9&&!a[--n]?[1,0]:x)

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Commented

n => s =>                       // given n and s

  (a = s.match(/10|./g))        // split s into marks; a '1' followed by a '0' is always

                                // interpreted as '10'

  .flatMap(x =>                 // for each mark x:

    x > 9 &&                    //   if x = '10',

    !a[--n] ?                   //   then decrement n; if a[n] is undefined:

      [1, 0]                    //     yield [1, 0]

    :                           //   else:

      x                         //     yield the mark unchanged

  )                             // end of flatMap()

JavaScript (ES6),  64  59 bytes

Saved 5 bytes thanks to @guest271314

Takes input as (n)(line).

n=>g=([...s])=>1/s[n]?g(eval(`[${s}]`.replace('1,0',10))):s

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Commented

n =>                            // main function, taking n

  g = ([...s]) =>               // g = recursive function, taking s

                                //     (which is either a string or an array)

    1 / s[n] ?                  // if s[n] is defined (i.e. we have too many marks):

      g(                        //   do a recursive call to g:

        eval(                   //     build a new array by evaluating ...

          `[${s}]`              //       ... the string representation of s where the

          .replace('1,0', 10)   //       first occurrence of '1,0' is replaced with '10'

        )                       //     end of eval()

      )                         //   end of recursive call

    :                           // else:

      s                         //   return s

Java (OpenJDK 8), 78 bytes

A nice one-liner using the streams API.

(n,l)->l.join(":",l.split("10",l.length()-n+1)).chars().map(i->i-48).toArray()

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How it works

(n,l)->                     // Lambda function taking int and string

  l.join(":",               // Join the following array with colons

    l.split("10",           // Split the original string on "10"...

      l.length()-n+1))      // But limit the parts to the difference between the length

                            // and expected length, to only remove  required number of 10s              

  .chars()                  // Convert to an intstream of codepoints

  .map(i->i-48)             // Remove 48 to get the numeric value of each codepoint

  .toArray()                // Return an int array

R, 63 bytes

While the length of the string is larger than n, substitute the next 10 you reach for a ":" (the ASCII character after 9). Then split into numbers by taking the ASCII value of each char in the string.

function(n,x){while(nchar(x)>n)x=sub(10,":",x);utf8ToInt(x)-48}

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Red, 91 bytes

func[n s][while[n < length? s][replace s"10""a"]foreach c s[prin[either c =#"a"[10][c]""]]]

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Jelly, 18 bytes

Ḍ⁵⁻ƊƝr1ŒpS‘⁼ɗƇḢk⁸Ḍ

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