A variable, $Y$, depends on $N$ independent variables, $n_1, n_2, n_3, ..., n_N$. This variable is extensive in the sense that:
$$Y(\lambda n_1,\lambda n_2,...,\lambda n_N)=\lambda Y(n_1,n_2,...,n_N) \tag{1}$$
where $n_j$ is the amount of component $j$. It can be proven that
$$Y=\Sigma_in_i\bar{Y}_i\tag{2}$$
where
$$\bar{Y}_i=(\frac{\partial{Y}}{\partial n_i})_{n_{j\neq i}} \tag{3}$$
In the proof showing (2) from (1), below is the first step:
We start by setting $v_i=\lambda n_i$, then differentiating (1) WRT $\lambda$ gives
$$\Sigma_{i=1}^{N}\frac{\partial{Y}}{\partial v_i}\cdot \frac{\partial{v_i}}{\partial \lambda}=\lambda Y(n_1,n_2,...,n_N) \tag{4}$$
I don't understand this step. Differentiating both sides of (1) WRT $\lambda$, we have
$$\frac{\partial}{\partial \lambda}Y(v_1,v_2,...v_N)=Y(n_1,n_2,...,n_N)+\lambda\frac{\partial{Y(n_1,n_2,...,n_N)}}{\partial{\lambda}}$$
And then sum over all $N$ components, shouldn't it be
$$\Sigma_{i=1}^{N}\frac{\partial{Y}}{\partial v_i}\cdot \frac{\partial{v_i}}{\partial \lambda}=NY(n_1,n_2,...,n_N)+N\lambda\frac{\partial{Y(n_1,n_2,...,n_N)}}{\partial{\lambda}}?$$
How do you get the RHS term in Eq(4)? I am just a lowly chemist, apologise if I asked something stupid.
One of the confusing things here is that you aren't giving a fixed name to the arguments of $Y$. Let's say the arguments of $Y$ are actually denoted by $x_i$ regardless of whether they are scaled or not.
Now differentiate (1) wrt $\lambda$. On the left side you initially get $\sum_{i=1}^n \frac{\partial Y}{\partial x_i}(\lambda n_1,\dots,\lambda n_N) \frac{\partial x_i}{\partial \lambda}$. Since $x_i=\lambda n_i$ here, you get $\sum_{i=1}^n n_i \frac{\partial Y}{\partial x_i}$.
On the other side, the derivative with respect to $\lambda$ is just $Y(n_1,\dots,n_N)$, because on that side $x_i$ is just $n_i$. So you have
$$\sum_{i=1}^n n_i \frac{\partial Y}{\partial x_i}(\lambda n_1,\dots,\lambda n_N) = Y(n_1,n_2,\dots,n_N).$$
The final step is to set $\lambda=1$ to get the desired result.