Let $$ R = \mathbb{Z}_m\times\cdots\times\mathbb{Z}_m = \times_{i=1}^\ell\mathbb{Z}_m $$ I'm trying to find the additive order of elements of $R$.
The additive order $\text{ord}(a)$ of $a\in\mathbb{Z}_m$ is the smallest integer $k\in\mathbb{N}$ such that $$ k\cdot a\equiv0\;\bmod m. $$ And resp. for $r\in R$ $$ \text{ord}(r)=\text{min}\{k\in\mathbb{N}:\forall i:k\cdot r_i\equiv0\;\bmod m\}. $$
So far I know that for $a\in\mathbb{Z}_m$ $$ \text{ord}(a) = \frac{\text{lcm}(a, m)}{a} $$ so I would expect $$ \text{ord}(r) = \text{lcm}\left(\frac{\text{lcm}(r_1, m)}{r_1},\ldots,\frac{\text{lcm}(r_\ell, m)}{r_\ell}\right) $$ for $r=(r_1,\ldots,r_\ell)\in R$.
But here I'm already stuck. Is there a way to simplify this any further?
Since you're only concerned with additive order, we can disregard the ring structure and the question becomes that is the order of an element from the direct product of groups. In that context, we have that the order of an element from the direct product of groups is the lcm of the order of each coordinate, which is exactly as you suggest. This can be seen fairly easily:
Suppose $g= (g_1,...,g_\ell)$ is an element of some direct product of $\ell$ groups. Let $L$ be the lcm of the orders of $g_1,...,g_\ell$. Then clearly $$ L(g_1,...,g_\ell) = (Lg_1,...,Lg_\ell) = 1 $$ So the order of $g$ is divisible by $L$. Suppose $g$ has order $k$ strictly less than $L$, then $kg_i = 1$ for all $g_1,...,g_\ell$, but for at least one such $g_i$, $k$ is not a multiple of its order so we have contraction. Thus $g$ has order $L$ and we can apply the result you invoked to get the order of the element in this specific case.
Also this is pretty, minor, but usually the direct product of rings, groups, etc. when you have a set of them is denoted as $$ \prod_{i= 1}^\ell\mathbb{Z}_m $$ instead of the $\times_{i = 1}^\ell \mathbb{Z}_m$, which may help you find answers to this kind of question more easily in the future.