But then what if the message $m_1+m_2$(or randomness $r_1+r_2$) overflows
For example messages are in field mod 2^64, than if message becomes some 2^64+1, it would becomes 1
With Pedersen, the points $G, H$ have a prime order (lets call that $q$); when you add the two commitments, you effectively get $G(m_1 + m_2 \bmod q) + H(r_1 + r_2 \bmod q)$. It doesn't matter how you thought about the messages before you generated the commitments. For example, you always picked the $m_1, m_2$ values from 0 to $2^{64}-1$ and think of them as values from $\mathbb{Z}_{2^{64}}$, Pederson will still add them modulo $q$.
BTW: you really do need to pick the $r_1, r_2$ values randomly from 0 to $q-1$ - otherwise, you lose the hiding property. For example, if you did select $r_1 \in [0, 2^{64}-1]$ (and the attacker knew that), then he could test whether a specific $m_1$ value was likely what was committed to with $O(2^{32})$ effort, which is quite practical.
But then what if the message $m_1+m_2$(or randomness $r_1+r_2$) overflows
Nothing special happens; the logical additions happen modulo $q$, no big deal; they'll wrap around. Of course, if you do select $m_1, m_2 \in [0, 2^{64}-1]$ (which is quite safe, unlike the case of $r_1, r_2$), that addition won't wrap (because $q \ggg 2^{64}$, at least, for any curve which is actually secure)
