No, the factorials grow too quickly to be diagonal coefficients of rational functions.

@PerAlexandersson I don't think so, but every root of unity is a primitive root of unity.

To give credit where credit is due, the $q$-Lucas theorem was first published, as far as I know, by Gloria Olive, Generalized powers, Amer. Math. Monthly 72 (1965), 619–625, equation (1.2.4). (The proof, which is not difficult, is apparently in her 1963 Ph.D. thesis, which I have not seen.) The theorem has been rediscovered many times since then, but I have not come across any earlier occurrence of it.

For some information on coefficients of the closely related analogous polynomials for Stirling numbers of the second kind, see my paper "On Miki's identity for Bernoulli numbers", J. Number Theory, 110 (2005), 75–82, people.brandeis.edu/~gessel/homepage/papers/miki2.pdf

What sort of formula are you looking for? There might be a formula as a double sum, but probably nothing simpler. Would that do any good?

I don't think you need anything more than Darij's first step. This gives the $(l+1)$th (or is it $(l+1)$st?) difference of a polynomial in $k$ of degree $m+1$

Incidentally, the sum can also be evaluated by Vandermonde's theorem, though this is overkill.

@Brendan Guilfoyle If $P(k)$ is a polynomial in $k$ of degree less than $l$ then $$\sum_{k=0}^l (-1)^k \binom lk P(k) = 0.$$ In your sum, we can write $\frac{k+2}{k+1}\binom{k+1}{m}$ as $P(k)= \frac{k+2}{m}\binom{k}{m-1}$, a polynomial in $k$ of degree $m$, and the sum $\sum_{k=m-1}^l$ can be replaced with $\sum_{k=0}^l$, since the additional terms are all zero. For a brief discussion of differences of polynomials, see artofproblemsolving.com/community/c6h90529. For a slightly more detailed discussion, see arxiv.org/abs/1609.05988, pp. 13--14.

The first case is easy as the sum is the $l$th difference of a polynomial of degree $m$.

It is possible to prove Abel identities automatically. See, e.g., Shalosh B. Ekhad and John E. Majewicz, A short WZ-style proof of Abel's identity, The Foata Festschrift, Electron. J. Combin. 3 (1996), no. 2, Research Paper 16, combinatorics.org/ojs/index.php/eljc/article/view/v3i2r16

To prove that $\sum_{i=0}^k a_i = b_k$ we need only check that $a_0=b_0$ and $b_k - b_{k-1}=a_k$.