Can we easily check if we can place two not-intersecting circles inside a convex polygon
Let $P$ be the given polygon. Let $Q$ be the set of points inside $P$ whose distance to the boundary of $P$ is at least $r$. Your requirement can be met iff the diameter of $Q$ is at least $2r$. Indeed a radius $r$ circle lies inside $P$ iff its center is in $Q$, and two such circles are disjoint iff the distance between their centers is at least $2r$.

The problem of finding $Q$ is called inward polygon offsetting. That link has a lot of discussion about the general case, but it should be much simpler when $P$ is convex.

Suppose the vertices of $P$ are $p_1,\ldots,p_n$ in counterclockwise order. Let $R$ denote rotation counterclockwise by $90^\circ$. Let $n_i$ be the outward unit normal vector to edge $p_ip_{i+1}$, namely $n_i=R(p_i-p_{i+1})/|R(p_i-p_{i+1})|$. Let $e_i$ denote this edge shifted by $r$, namely the line from $p_i-rn_i$ to $p_{i+1}-rn_i$. Now we can iteratively construct a path $q$ which will become the boundary of $q$. In iteration $1$, let $q=e_1$. In iteration $i$, find the intersection of $q$ with $e_i$, discard the part of $q$ after the intersection and append the part of $e_i$ after the intersection (we should look for the intersection starting from the end of $q$; the more segments we have to check, the more vertices we can discard, so the algorithm will still be $O(n)$). After $n$ iterations, find where $q$ intersects itself and discard excess vertices to obtain a closed path; this is the boundary of $Q$.

Once we have $Q$, we can find its diameter using the Rotating calipers algorithm.