This is a follow-up to a much simpler question I asked here, which @PrincessEev answered promptly and perfectly. She showed me how to rewrite the sum $\sum _{i=1}^x \phi (x-i)$ in such a way that Abel Summation could be applied, to obtain
$$\sum _{i=1}^x \phi (x-i) = \lfloor x \rfloor \phi(x-1) - \int_0^{x-1} \lfloor u+1 \rfloor \phi'(u) \, \mathrm{d}u$$
Here is a harder problem - I think. And once again, I don't really know where to start. Let $1 \le j \le x$ be a constant positive integer. Can one apply Abel Summation to the sum $\sum _{i=1}^x \phi (x - i j)$? If so, how?
CLARIFICATION / UPDATE
I initially added follow-up material here asking what I was doing wrong in implementing @junjios's answer. The fault was mine and was a coding error in my use of Mathematica.
Since it has no bearing on @junjios's correct answer, I have deleted those elements of this post, to avoid confusing new readers. If you are puzzled over some of the posts in the comments section, this is the explanation.
Let $x\geq 1$ and $j\in\mathbb{R}$ be fixed. (Note: There are no further conditions on $j$!)
We define: $\psi(y):=\phi(x-yj)$. Clearly, $\psi$ is continuously differentiable, so we can apply the Abel summation formula to obtain $$\sum_{i=1}^x\phi(x-ij)=\sum_{i=1}^x\psi(i)=\lfloor x\rfloor\psi(x)-\int_1^x\lfloor u\rfloor\psi'(u)\,\mathrm{d}u\\=\lfloor x\rfloor\phi(x(1-j))+j\int_1^x\lfloor u\rfloor\phi'(x-uj)\,\mathrm{d}u.$$
Generalisation:
More generally, given any (further) continuously differentiable function $f$ (possibly depending on $x$), we may write $$\sum_{i=1}^x\phi(f(i))=\lfloor x\rfloor\phi(f(x))-\int_1^x\lfloor u\rfloor\phi'(f(u))f'(u)\,\mathrm{d}u.$$