Take a fixed $m$ and first solve the recursion relation for $f(k,m)$, $$ f(k,m)= \frac{m^{k+1}-1}{m-1},\;\;m\neq 1, $$ with $\lim_{m\rightarrow 1} f(k,m)=k+1$.
Then define the generating function $$ G_n(z)=\sum_{k\ge 0} R(n,k,m)z^k. $$ The recursion relation and initial conditions give \begin{eqnarray*} &(1-z)G_n(z)=nG_{n-1}(z)+G_{n-1}'(z),\\ &G_0(z)=\sum_{k\ge 0} f(k,m)z^k=\frac{1}{(1-z)(1-mz)}. \end{eqnarray*} We introduce the exponential generating function $$H(x,z)=\sum_{n\geq 0}G_n(z)\frac{x^n}{n!},$$ which satisfies the differential equation $$(1-x-z)\frac{\partial H}{\partial x}-\frac{\partial H}{\partial z}=H(x,z).$$ The initial condition is $H(0,z)=G_0(z)$.
Mathematica solves the differential equation for $H(x,z)$ in terms of the (principal branch) Lambert W-function, which gives $$H(x,0)=\frac{W(-x)}{-x[W(-x)+1] [m W(-x)+1]},\;\;x\leq 1.$$ Since $$H(x,0)=\sum_{n\geq 0}R(n,0,m)\frac{x^n}{n!}, $$ we need to extract the term of order $x^n$ in the expansion of $H(x,0)$ in powers of $x$. This is done below. The result is \begin{eqnarray*} H(x,0)={}&\sum_{n\geq 0}\left(\sum_{r=1}^{n+1} \frac{m^{r-1}n!(n+1)^{n+1-r}}{(n+1-r)!}\right)\frac{x^n}{n!}. \end{eqnarray*} The sum over $r$ is $T(n+1,m)$, so we have proven the desired result $$R(n,0,m)=T(n+1,m).$$
Calculation of the series expansion of $H(x,0)$.
The Lambert W-function produces the expansions \begin{eqnarray*} &W(-x)^r=(-1)^r r\sum_{n=r}^\infty \frac{n^{n-r-1}}{(n-r)!}x^n,\\ &\frac{1}{W(-x)+1}=\sum_{n=0}^\infty\frac{n^n}{n!}x^n, \end{eqnarray*} which we substitute into \begin{eqnarray*} H(x,0)={}&\frac{1}{-x}\frac{1}{W(-x)+1}\sum_{r=1}^\infty (-m)^{r-1} W(-x)^{r}\\ ={}&\frac{1}{x}\left(\sum_{n=0}^\infty\frac{n^n}{n!}x^n\right)\sum_{r=1}^\infty m^{r-1}\left( r\sum_{n=r}^\infty \frac{n^{n-r-1}}{(n-r)!}x^n\right). \end{eqnarray*} Collecting powers of $x$, one arrives at \begin{eqnarray*} H(x,0)={}&\sum_{n=0}^{\infty}\left(\sum_{r=1}^{n+1}m^{r-1} s_{n,r}\right)\frac{x^n}{n!},\\ s_{n,r}={}&rn!\sum_{k=0}^{n+1-r}\frac{k^k}{k!}\frac{(n+1-k)^{n-k-r}}{(n+1-k-r)!}. \end{eqnarray*} It remains to evaluate $s_{n,r}$, which can be done with the help of Abel's binomial theorem. (Thanks to ChatGPT for pointing me to that classic result.) First rewrite the sum identically in terms of binomial coefficients. Define $N_r=n+1-r$,
$$s_{n,r}=\frac{(N_r+r-1)!}{N_r!}r\sum_{j=0}^{N_r}{{N_r}\choose j}(N_r-j)^{N_r-j}(r+j)^{j-1}. $$ Abel's formula gives $$r\sum_{j=0}^{N_r}{{N_r}\choose j}(N_r-j)^{N_r-j}(r+j)^{j-1}=(N_r+r)^{N_r}.$$ We thus arrive at $$s_{n,r}=\frac{n!}{(n+1-r)!}(n+1)^{n+1-r}.$$
