They are the exact same, and there isn't really a reason why one is "better" than the other, though I will talk about that at the end.
To see that the definitions are equivalent, we simply have to show that for all $n \geq 1$, $\mu_V(n) = \mu_W(n)$ where $\mu_V$ is by Vinogradov and $\mu_W$ is by um Wikipedia. Suppose that $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ where $e_i \geq 1$ and $p_i$ are distinct. Then, by definition of multiplicative function and the fact that $p_i$ distinct implies $p_i^{e_i}$ coprime, we have
\begin{align*}
\mu_V(n)
&= \mu(p_1^{e_1}) \mu(p_2^{e_2}) \cdots \mu(p_k^{e_k}) \\
&= \left(\begin{cases} -1 &\text{if}\, e_1 = 1 \\ 0 &\text{otherwise} \end{cases}\right) \cdot \left(\begin{cases} -1 &\text{if}\, e_2 = 1 \\ 0 &\text{otherwise} \end{cases}\right) \cdots \left(\begin{cases} -1 &\text{if}\, e_k = 1 \\ 0 &\text{otherwise} \end{cases}\right) \\
&= \begin{cases} (-1)^k &\text{if}\, e_1 = e_2 = \cdots = e_k = 1 \\ 0 &\text{otherwise} \end{cases}
\end{align*}
Which you see is exactly the same as $\mu_W$, since the condition $e_1 = e_2 = \cdots = e_k = 1$ is the definition of square-free, and simply do casework on $k$ mod $2$.
I think both definitions are useful in their own way. On one hand, Vinogradov's definition is short and concise, and is useful for people with decent familiarity with the topic and just to clearify notation. It is also "minimal" information required to define an arithmetic function, since $f(p^k)$ defines an arithmetic function $f$. On the other hand, the wikipedia definition is more useful for people seeing it the first time (which is what wikipedia is for). It is also more natural if you motivate the definition of mobius function from inclusion-exclusion principle.
Just to elaborate on the IEP, consider the sieve of eratosthenes in the following formulation. Let $A = \{1, 2, \ldots, N\}$ and let $A_d = \{k \in A: d \mid k\}$. Then number of primes under $N$ is then $\left|A - \bigcup_{p \in \mathcal{P}} A_p\right| = N - \left|\bigcup_{p \in \mathcal{P}} A_p\right|$. If we write the union out using IEP and let $f(d) = |A_d| = \lfloor\frac{N}{d}\rfloor$ we get
\begin{align*}
N - \left|\bigcup_{p \in \mathcal{P}} A_p \right|
&= f(1) - \left(f(2) + f(3) + f(5) + \cdots\right) \\
&+ \left(f(6) + f(10) + f(15) + \cdots + f(p_1p_2)\right) \\
&- \sum_{p_1<p_2<p_3\in\mathcal{P}} f(p_1p_2p_3) + \cdots \\
&= f(1) - f(2) - f(3) - f(5) + f(6) - f(7) + f(10) - f(11) + \cdots \\
&:= \sum_{n=1}^N \mu(n)f(n)
\end{align*}
Cheers!