Convolution and DFT size
Note that if $x[n]$ and $y[n]$ are non-zero in the interval $n\in [0,2N-1]$ (and, for convenience, they are defined as being zero outside that interval), the (linear) convolution sum can be written as

$$y[n]=(x\star h)[n]=\sum_{m=0}^nx[m]h[n-m],\qquad 0\le n&lt;4N-1\tag{1}$$

Note that the upper limit is $n$ because the argument of the term $h[n-m]$ becomes negative for $m&gt;n$.

So if, for whatever reason, you're only interested in $y[n]$ in the interval $n\in [0,N-1]$, you can set the upper limit in the convolution sum $(1)$ to $N-1$ (assuming that $x[n]$ and $h[n]$ are defined as zero outside the interval $n\in [0,2N-1]$). However, the result wouldn't change if you chose the upper summation interval as $2N-1$, because you would just add zeros to the result.

In short, both formulas in your question give the same result, under the given assumptions on the definitions of $x[n]$ and $y[n]$.