In the body of this challenge, \$\begin{pmatrix}n\\k\end{pmatrix}\$ is used to represent the number of combinations of \$k\$ elements of \$n\$, also written as \$\frac{n!}{k!(n-k)!}\$ or \$n\mathrm{C}r\$.
Any nonnegative integer \$m\$, for arbitrary natural (positive) \$r\$, can be written as a unique series of \$r\$ combinations such that $$m=\sum\limits_{i=1}^{r}\begin{pmatrix}C_i\\i\end{pmatrix}$$ provided the sequence \$C\$ both strictly increases (i.e. \$C_{\ell-1}\lneq C_\ell\$) and consists solely of nonnegative integers. \$C\$ is not necessarily unique without these restrictions.
Example
Consider \$m=19\$ and \$r=4\$. Values of \$C_4\$, \$C_3\$, \$C_2\$ and \$C_1\$ must be found for the equation $$19=\sum\limits_{i=1}^4\begin{pmatrix}C_i\\i\end{pmatrix}\\$$ which can be rewritten as $$\begin{pmatrix}C_4\\4\end{pmatrix}+\begin{pmatrix}C_3\\3\end{pmatrix}+\begin{pmatrix}C_2\\2\end{pmatrix}+\begin{pmatrix}C_1\\1\end{pmatrix}=19$$ Begin by finding the largest value of \$C_4\$ which satisfies the inequality \$\begin{pmatrix}C_4\\4\end{pmatrix}\leq 19\$. \$C_4\$ is six: $$\begin{pmatrix}6\\4\end{pmatrix}+\begin{pmatrix}C_3\\3\end{pmatrix}+\begin{pmatrix}C_2\\2\end{pmatrix}+\begin{pmatrix}C_1\\1\end{pmatrix}=19\\15+\begin{pmatrix}C_3\\3\end{pmatrix}+\begin{pmatrix}C_2\\2\end{pmatrix}+\begin{pmatrix}C_1\\1\end{pmatrix}=19\\\begin{pmatrix}C_3\\3\end{pmatrix}+\begin{pmatrix}C_2\\2\end{pmatrix}+\begin{pmatrix}C_1\\1\end{pmatrix}=4$$ The problem has been reduced to \$m=4\$ and \$r=3\$. The largest value of \$C_3\$ which satisfies the inequalities \$\begin{pmatrix}C_3\\3\end{pmatrix}\leq4\$ and \$C_3\lneq C_4\$ must be found. \$C_3\$ is four: $$\begin{pmatrix}4\\3\end{pmatrix}+\begin{pmatrix}C_2\\2\end{pmatrix}+\begin{pmatrix}C_1\\1\end{pmatrix}=4\\4+\begin{pmatrix}C_2\\2\end{pmatrix}+\begin{pmatrix}C_1\\1\end{pmatrix}=4\\\begin{pmatrix}C_2\\2\end{pmatrix}+\begin{pmatrix}C_1\\1\end{pmatrix}=0$$ Any combination of the form \$\begin{pmatrix}n\\k\end{pmatrix}\$ with \$n<k\$ is zero, and so \$C_2=1\$ and \$C_1=0\$: $$\begin{pmatrix}1\\2\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}=0\\0+0=0\\0=0\checkmark$$
Note that \$C_2\$ cannot be zero because then \$C\$ would not strictly increase unless \$C_1\$ were negative, which cannot be the case due to the condition that \$C\$ consists solely of nonnegative integers. The solution is summarized with the statement \$C=(0,1,4,6)\$ (here, 1-based indexing is used). The process followed here is guaranteed to produce the correct \$C\$.
The Challenge
Given \$m\$ and \$r\$, find the elements of \$C\$.
Rules
This is code-golf so the shortest answer in bytes wins.
Assume only valid input will be given.
Input and output may assume whatever form is most convenient. This can include outputting the elements of \$C\$ in any order, because \$C\$ strictly increases and so the actual order of the elements is trivially found by sorting them.
Terms whose combinations evaluate to zero, e.g. \$C_2\$ and \$C_1\$ in the example, cannot be neglected in output.
A program should theoretically work for arbitrarily large values of \$m\$ and \$r\$, but is still acceptable if it is limited by memory constraints.
Test Cases
Here \$m\$ is the first number and \$r\$ is the second, and the output begins with \$C_1\$.
In: 19 4 Out: 0 1 4 6 In: 0 4 Out: 0 1 2 3 In: 40000 6 Out: 6 8 9 11 12 20 In: 6 6 Out: 1 2 3 4 5 6 In: 6 5 Out: 0 1 2 3 6 In: 6 20 Out: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 15 16 17 18 19 20 (note 14 is skipped) In: 6 1 Out: 6 
m=0, my mistake. You are correct that r is only ever positive (edited to note this) @JonathanAllan \$\endgroup\$