The methods are not equivalent. To see this, let us pick $N=2$, denote the random probabilities by $(P_1,P_2)$ and compute
\begin{equation}
\mathbb{P}\left(P_1 \leq \frac{2}{3}\right).
\end{equation}
for both methods.
Method 1
Method 1 in the $N=2$ case works as follows. $X_1,X_2$ are iid. $\mathrm{Uniform}(0,1)$ random variables and
\begin{equation}
P_1 = \frac{X_1}{X_1 + X_2}
\end{equation}
Therefore,
\begin{equation}
\mathbb{P}\left(P_1 \leq \frac{2}{3}\right) = \mathbb{P}\left(\frac{X_1}{X_1+X_2} \right) = \mathbb{P}\left(X_1 \leq 2X_2 \right).
\end{equation}
There are various straightforward ways to evaluate the last expression. For example, $(X_1,X_2)$ is uniform in the unit square $[0,1]\times[0,1]$, so the probability is simply the area where $X_1\leq X_2$. Alternatively, let us decompose it into two conditional probabilities conditioning on whether $X_2\leq\frac{1}{2}$:
\begin{equation}
=\mathbb{P}(X_2\leq \frac{1}{2}) \mathbb{P}\left(X_1 \leq 2X_2 \mid X_2 \leq \frac{1}{2}\right) + \mathbb{P}(X_2 > \frac{1}{2}) \mathbb{P}\left(X_1 \leq 2X_2 \mid X_2 > \frac{1}{2}\right)
\end{equation}.
Conditional on $X_2 \leq\frac{1}{2}$, $2X_2$ is uniform in $[0,1]$ and thus for symmetry reasons $\mathbb{P}\left(X_1 \leq 2X_2 \mid X_2 \leq \frac{1}{2}\right) = \frac{1}{2}$. If $X_2 > \frac{1}{2}$, $2X_2 > 1$ and therefore $X_1 < X_2$. Thus $\mathbb{P}\left(X_1 \leq 2X_2 \mid X_2 > \frac{1}{2}\right) = 1$. Finally, $\mathbb{P}(X_2\leq \frac{1}{2}) = \mathbb{P}(X_2 > \frac{1}{2}) = \frac{1}{2}$. Plugging these values into the expression, we get
\begin{equation}
= \frac{1}{2}\times \frac{1}{2} + \frac{1}{2} \times 1 = \frac{3}{4}.
\end{equation}
Method 2
In the $N=2$ case there is only one cutpoint uniform in $[0,1]$ and this cutpoint directly equals $P_1$. Thus $P_1 \sim \mathrm{Uniform}(0,1)$ and
\begin{equation}
\mathbb{P}\left(P_1 \leq \frac{2}{3} \right) = \frac{2}{3},
\end{equation}
which is not equal to the value obtained with Method 1.
Which one to use
As kjetil b halvorsen pointed out in comments, it is impossible to judge which is better for your use case without knowing more details about your use case. Besides these approaches, you may also want to investigate Dirichlet distribution which is often used in Bayesian statistics to define probability distributions over discrete probability distributions (just as the methods described in the question). The Dirichlet distribution has $N$ parameters which you could tune to obtain distributions closer to desired behavior. On the other hand, the use of the Dirichlet distribution is mainly motivated by its useful computational properties in the Bayesian context, which may be totally irrelevant in your use case.
Another suggestion is to do your simulations using various distributions over the probabilities. If the conclusions you are interested in do not change, good, if they change, you need to think more about which distribution you should use.
