You have a product of shape
$$ \sum_{k = 0}^\infty a_k \sum_{\ell = 0}^n b_\ell. \tag{1}$$
You will notice that I've done a few things differently than you. Firstly, I've given the two (somethings) names. Secondly, and more importantly, I've indicated that the first sum is over $k$ and the second sum is over $\ell$. It is important to not reuse indices (like $k$) for different sums in the same expression in order to retain well-definedness.
The product is not exciting. In short, the product is written in the exact same way as one might think of
$$ (a_1 + a_2 + a_3)(b_1 + b_2 + b_3) = a_1b_1 + a_1b_2 + a_1b_3 + \cdots + a_3b_1 + a_3b_2 + a_3b_3,$$
in that the product is each possible choice of first multiplicand and second multiplicand, added together.
One way to write $(1)$ is to write
$$ \sum_{k = 0}^\infty \sum_{\ell = 0}^n a_k b_\ell.$$
This is not exciting, but it is the product; and it is exactly the sum over choices of first and second multiplicand.
I suspect you mean something a little bit different. So let's now think about sums of shape
$$ \sum_{k = 0}^\infty a_k x^k \sum_{\ell = 0}^n b_\ell x^\ell.$$
It is a very reasonable question to wonder what the coefficient of $x^m$ is in this product. So let's first rewrite the product,
$$ \sum_{k = 0}^\infty \sum_{\ell = 0}^n a_k b_\ell x^{k + \ell}.\tag{2}$$
The coefficient of $x^m$ comes from those terms with an $x^m$ appearing, which happen when $k + \ell = m$.
When can $k + \ell = m$? you are summing over all natural numbers $k$ and those natural numbers $\ell$ up to $n$. So for $m \leq n$, we can think of pairing $\ell$ with $k = m - \ell$ to see that the coefficient of $x^m$ in $(2)$ is
$$ \sum_{k + \ell = m} a_k b_\ell = \sum_{\ell = 0}^m a_{m - \ell} b_{\ell}.$$
For $m > n$, we only have $n$ terms. We can still pair $\ell$ with $k = m - \ell$, but only for $\ell$ up to $n$. So for $m > n$, the coefficient of $x^m$ in $(2)$ is
$$ \sum_{k + \ell = m}' a_k b_\ell = \sum_{\ell = 0}^n a_{m - \ell}b_\ell.$$
Aside: You will notice that I've used a primed summation symbol --- this is to indicate that there is something strange about the sum. Here, I mean that there is a restriction on $\ell$ which I did not make explicit in the first sum. Sometimes, people are more explicit, and they might write it as
$$ \sum_{\substack{k + \ell = m \\ 0 \leq \ell \leq n}} a_k b_\ell.$$
But if you have never seen conditions below indices of summation (like I expect that the OP has not seen these before), this can look intimidating.
So you could regroup all of these together and write the sum as
$$ \sum_{m = 0}^n \sum_{\ell = 0}^m a_{m - \ell} b_\ell x^m + \sum_{m = n+1}^\infty \sum_{\ell = 0}^n a_{m - \ell} b_\ell x^m.$$
Or, more compactly but less legibly, you could write it as
$$ \sum_{m = 0}^\infty \sum_{\ell = 0}^{\min\{m,n\}} a_{m - \ell} b_\ell x^m,$$
where $\min\{m, n\}$ denotes the minimum of $m$ and $n$.
