I'll show the answer is $\frac{1}{2}\log(p)+O(1)$ when $k = \frac{p}{2}$, and $\frac{5}{6}\log(p)+O(1)$ when $k = \frac{p}{3}$.
For ease, I'll assume $p$ is prime. I'll also ignore $a=0$, which obviously is allowed.
Let's first greatly simplify the problem. Since $a \mapsto -a^{-1}$ is a bijection, the expectation is equivalent to
\begin{align*}
\mathbb{E}_{a \in \mathbb{Z}_p} \min(-a^{-1},-2a^{-1},\dots,-ka^{-1}) &= \mathbb{E}_{a \in \mathbb{Z}_p} \min\{j \ge 1 : j \equiv -a^{-1}x \text{ for some } x \in [k]\} \\ &= \mathbb{E}_{a \in \mathbb{Z}_p} \min\{j \ge 1 : ja \equiv -x \text{ for some } x \in [k]\} \\ &= \mathbb{E}_{a \in \mathbb{Z}_p} \min\{j \ge 1 : ja \in \{p-k,\dots,p-1\}\}.
\end{align*}
So, if $k = \frac{p-1}{2}$, we obtain
\begin{align*}
\mathbb{E}_{a \in \mathbb{Z}_p} \min\{j \ge 1 : ja \in \{\frac{p+1}{2},\dots,p-1\}\} &= \frac{1}{p-1}\left[\sum_{a=1}^{(p-1)/2} \lceil \frac{(p+1)/2}{a} \rceil + \sum_{a=(p+1)/2}^{p-1} 1\right] \\ &= \frac{1}{p-1}\left[\sum_{a=1}^{(p-1)/2} [\frac{(p+1)/2}{a} + O(1)] + \frac{p}{2}+O(1)\right] \\& = \frac{1}{2}\log(p)+O(1).
\end{align*}
Now take $k = \frac{p-1}{3}$, or $\frac{p-2}{3}$ if $p \equiv 2 \pmod{3}$ -- I'll just write $p/3$ for ease.
If $a \le \frac{p}{3}$, then since the "jumps" between $ja$ and $(j+1)a$ are at most $\frac{p}{3}$, we have $\min\{j : ja \in [\frac{2p}{3},p]\} = \lceil \frac{2p/3}{a} \rceil$. If $\frac{p}{3} \le a \le \frac{p}{2}$, then $2a \in [\frac{2p}{3},p]$. If $\frac{2p}{3} \le a \le p$, then $a \in [\frac{2p}{3},p]$. Take $a \in [\frac{p}{2},\frac{2p}{3}]$, and let $\Delta = a-\frac{p}{2}$. Then, for $j$ even, $ja \equiv j\Delta$, and for $j$ odd, $ja \equiv \frac{p}{2}+j\Delta$. Since $\Delta \le \frac{p}{6}$, $2\Delta \le \frac{p}{3}$, and so it can't be the case that $j\Delta$ to $(j+2)\Delta$ jumps over $[\frac{2p}{3},p]$. We therefore see that $\min\{j : ja \in [\frac{2p}{3},p]\} = \lceil \frac{p/6}{\Delta}\rceil$. Putting everything together, $$\mathbb{E}_{a \in \mathbb{Z}_p} \min\{j \ge 1 : ja \in [\frac{2p}{3},p]\} = \frac{1}{p}\left[\sum_{a=1}^{p/3} \lceil \frac{2p/3}{a} \rceil + \sum_{a=p/3}^{p/2} 2 + \sum_{a=p/2}^{2p/3} \lceil \frac{p/6}{a-p/2}\rceil + \sum_{a=2p/3}^p 1\right],$$
which gives $\frac{5}{6}\log(p)+O(1)$.
The analysis done above for $k = p/3$ becomes (seemingly) much more complicated for $k = p/N$ for larger $N$ (though $k=p/4,p/5$ might be easily doable). Definitely, though, the answer will be $c_N \log(p)+O(1)$ for some constant $c_N$ depending only on $N$.
The situation becomes complicated when $p$ is composite. I'll write $n$ instead of $p$ for psychological ease. Suppose $n$ is even, and set $k = \frac{n}{2}$. Take $a \in [n]$. If $g := \gcd(a,n)$ is bigger than $1$, then $a\frac{n}{g} \equiv 0 \pmod{n}$ and thus the min is $0$. The set of $a \in \mathbb{Z}_n$ with $\gcd(a,n) = 1$ is permuted under the (well-defined) operation $a \mapsto -a^{-1}$, and so, as done when $n$ was prime, we get
\begin{align*}
\mathbb{E}_{a \in \mathbb{Z}_n} \min(a,2a,\dots,\frac{n}{2}a) &= \frac{1}{n}\sum_{\substack{a \in \mathbb{Z}_n \\ (a,n) = 1}} \min(-a^{-1},-2a^{-1},\dots,-\frac{n}{2}a^{-1}) \\ &= \frac{1}{n}\sum_{\substack{a \in \mathbb{Z}_n \\ (a,n) = 1}} \min\{j \ge 1 : aj \in \{\frac{n}{2}+1,\dots,n-1\}\} \\ &= \frac{1}{n}\left[\sum_{\substack{a = 1 \\ (a,n) = 1}}^{n/2} \lceil \frac{n/2}{a} \rceil + \sum_{\substack{a = n/2 \\ (a,n) = 1}} 1 \right].
\end{align*}
Now,
\begin{align*}
\sum_{\substack{a=1 \\ (a,n) = 1}}^{n/2} \frac{1}{a} &= \sum_{a=1}^{n/2} \frac{1}{a}\sum_{d \mid (a,n)} \mu(d) = \sum_{a=1}^{n/2}\frac{1}{a}\sum_{\substack{d \mid a \\ d \mid n}} \mu(d) \\ &= \sum_{d \mid n} \mu(d)\sum_{d \mid a \le \frac{n}{2}} \frac{1}{a} = \sum_{d \mid n} \mu(d)\sum_{k \le \frac{n}{2d}} \frac{1}{dk} \\ &= \sum_{d \mid n} \frac{\mu(d)}{d}\left[\log(\frac{n}{2d})+\gamma+O(\frac{2d}{n})\right].
\end{align*}
We have $$\sum_{d \mid n} \frac{\mu(d)}{d}O(\frac{2d}{n}) = \sum_{d \mid n} O(\frac{1}{n}) = O(1)$$ $$\sum_{d \mid n} \frac{\mu(d)}{d}\gamma = \gamma\frac{\phi(n)}{n} = O(1)$$ $$\sum_{d \mid n} \frac{\mu(d)}{d}\log(\frac{n}{2d}) = \sum_{d \mid n} \frac{\mu(d)}{d}\log(\frac{n}{d})-\log(2)\frac{\phi(n)}{n}.$$ Therefore, $$\mathbb{E}_{a \in \mathbb{Z}_n} \min(a,2a,\dots,\frac{n}{2}a) = \frac{1}{2}\sum_{d \mid n} \frac{\mu(d)}{d}\log(n/d)+O(1).$$ There might be a closed form for this quantity, but I can't figure it out right now. It definitely varies highly based on the factorization of $n$. If $n$ is prime, it is asymptotic to $\frac{1}{2}\log(n)$, while if $n = 2p$ for a prime $p$, it is asymptotic to $\frac{1}{4}\log(n)$.
ADDED: See here. It implies $$\frac{1}{2}\sum_{d \mid n} \frac{\mu(d)}{d}\log(n/d)+O(1) = \frac{1}{2}\log(n)\sum_{d \mid n} \frac{\mu(d)}{d} +O(\log\log n) = \frac{1}{2}\log(n)\frac{\phi(n)}{n},$$ giving the result $$\mathbb{E}_{a \in \mathbb{Z}_n} \min(a,2a,\dots,\frac{n}{2}a) = \frac{1}{2}\log(n)\frac{\phi(n)}{n}+O(\log\log n).$$
