J, 9 bytes

%&!*-![+-

Computes the same thing as everyone else. n as left arg, k as right arg.

%&!*-![+-
        -   NB. n-k
      [     NB. n
       +    NB. addition
    -       NB. n-k
     !      NB. combinations, x!y -> C(y,x)
%&!         NB. take factorial of n and k and divide the results
   *        NB. multiplication

More simply put

%&!*-![+-
    -![+-   NB. (n-k)!(n+n-k)
%&!         NB. (n!)%(k!)
   *        NB. multiply the results

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Vyxal, 9 8 bytes

2ʀ*Ṙ/

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Getting the inputs to work nicely was annoying, and it's likely that there's a byte to be shaved here (I wish vysearch was a thing). Same formula as everyone else. -1 inspired by Kevin Cruijssen's 05AB1E answer.

2ʀ       # [0, 1, 2]
  *Ṙ     # [2n, n, 0]
    ε    # [2n-k, n-k, k]
     ¡   # Factorial of each
      ƒ/ # (2n-k)! / (n-k)! / k!

05AB1E, 9 bytes

xs0)α!.»÷

Inputs in the order \$n,k\$.

Try it online or verify all test cases.

Explanation:

x          # Double the first (implicit) input (without popping)
           #  STACK: n,2n
 s         # Swap the two values
           #  STACK: 2n,n
  0        # Push 0
           #  STACK: 2n,n,0
   )       # Wrap all values on the stack into a list
           #  STACK: [2n,n,0]
    α      # Get the absolute difference with the second (implicit) input
           #  STACK: [abs(k-2n),abs(k-n),abs(k-0)], aka [2n-k,n-k,k]
     !     # Get the factorial of each
           #  STACK: [(2n-k)!,(n-k)!,k!]
      .»   # Left-reduce this list by:
        ÷  #  Integer-dividing
           #   STACK: (2n-k)!/(n-k)!/k!
           # (which is output implicitly as result)  

APL(Dyalog Unicode), 12 9 bytes ^SBCS

It takes n and k as its left and right arguments, respectively, then outputs the coefficient of \$x^k\$ in the \$n\$-th approximant. It calculates \$\frac{n!}{k!}\cdot{2n-k\choose n-k} = \frac{n!}{k!}\cdot\frac{(2n-k)!}{n!(n-k)!} = \frac{(2n-k)!}{k!(n-k)!}\$.

÷⍥!×-!⊣+-

÷⍥!        ⍝ n!÷k!
   ×       ⍝      times
    -      ⍝           n-k
     !     ⍝              choose from
      ⊣+-  ⍝                         n+n-k

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Try it on APLgolf! 12 bytes

JavaScript (ES6), 44 bytes

Expects (n)(k) and returns the corresponding coefficient.

n=>k=>(g=n=>n?n*g(n-1):1)(2*n-k)/g(k)/g(n-k)

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Ruby, 41 bytes

->n,k{eval [[1,*k...n+n-=k]*?*,*1..n]*?/}

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Same formula as @Tbw, but 1-based to make it shorter.

Uiua 0.13.0-dev.2, 15 bytes ^SBCS

÷∩/×+1⊃⇡-:+1+⇡.

Try on Uiua Pad!

Takes \$n\$ and \$k\$ and outputs the coefficient of \$x^k\$ in the numerator of the \$n\$th approximant. This is \$\frac{(2n-k)!}{k!(n-k)!}\$.

Charcoal, 18 bytes

ＮθＮηＩ÷Π⁻⊕…θ⊗θηΠ⊕…η

Attempt This Online! Link is to verbose version of code. Explanation: Uses the same formula from @Tbw's answer.

Ｎθ                  Input `n` as a number
  Ｎη                Input `k` as a number
         …          Range from
          θ         Input `n` to
            θ       Input `n`
           ⊗        Doubled
        ⊕           Vectorised incremented
       ⁻            Vectorised subtract
             η      Input `k`
      Π             Take the product, i.e. `(2n-k)!/(n-k)!`
     ÷              (Integer) divided by
                 η  Input `k`
                …   Range from `0`
               ⊕    Vectorised incremented
              Π     Take the product, i.e. `k!`
    Ｉ               Cast to string
                    Implicitly print