It is not hard to show that $0 \leq \delta_{k-1}(n)-\delta_{k-1}(n-1) \leq k-1$ and to say when each possible jump happens.
Let me first alter the notation by setting $a_j=j+t_j.$
For any two integers $k,n\ge 1$, there is a unique way of writing
$$n=\binom{k+t_k}{k}+\binom{k-1+t_{k-1}}{k-1}+\dots+\binom{i+t_i}{i}$$
so that $i \geq 1$ and $t_k\geq t_{k-1} \geq \dots \geq t_i\geq 0$.
Define
$$\partial_{k-1}(n) = \binom{k+t_k}{k-1}+\binom{k-1+t_{k-1}}{k-2}+\dots+\binom{i+t_i}{i-1}.$$
CLAIM: Let $n=\binom{k+t_k}{k}+\binom{k-1+t_{k-1}}{k-1}+\dots+\binom{i+t_i}{i}>1$ as above.
If $t_i = 0$ then
$n=\binom{k+t_k}{k}+\binom{k-1+t_{k-1}}{k-1}+\dots+\binom{i+1+t_{i+1}}{i+1}+\binom{i}{i}$
$n-1=\binom{k+t_k}{k}+\binom{k-1+t_{k-1}}{k-1}+\dots+\binom{i+1+t_{i+1}}{i+1}$
$\delta_{k-1}(n)-\delta_{k-1}(n-1)=i$.
If $t_i>0$ then
$n=\binom{k+t_k}{k}+\binom{k-1+t_{k-1}}{k-1}+\dots+\binom{i+1+t_{i+1}}{i+1}+\binom{i+t_i}{i}$
$n-1=\binom{k+t_k}{k}+\binom{k-1+t_{k-1}}{k-1}+\dots+\binom{i+1+t_{i+1}}{i+1}+\binom{i+t_i-1}{i}+\binom{i-1+t_i-1}{i-1}+\cdots+\binom{1+t_i-1}{1}$
$\delta_{k-1}(n)-\delta_{k-1}(n-1)=0$.
Note that if the expansion above for $n$ is according to the requirements then the expansion claimed for $n-1$ also meets the requirement. That it is actually equal to $n-1$ is clear in the case $t_i=0$ and for the case $t_i \gt 0$ follows from this familiar fact:
For $a \gt i$ and $i \geq 1$ we have $$\binom{a}{i}=\binom{a-1}{i}+\binom{a-2}{i-1}+\cdots+\binom{a-i}{1}+\binom{a-i-1}{0}$$ Switching to the $t$ notation:$$\binom{i+t}{i}=\binom{i+t-1}{i}+\binom{i-1+t-1}{i-1}+\cdots+\binom{1+t-1}{1}+1$$
That $\delta_{k-1}(n)-\delta_{k-1}(n-1)=i$ when $t_i=0$ is clear and that $\delta_{k-1}(n)=\delta_{k-1}(n-1)$ when $t_i \gt 0$ follows from writing the same fact as
$$\binom{i+t}{i-1}=\binom{i+t-1}{i-2}+\binom{i-1+t-1}{i-2}+\cdots+\binom{1+t-1}{0}$$