I can't see any match with the initial $y_n$ in equation $(1)$, (see update). With a slight change to the notation, you have the following results in the first two cases, for Notation 1 and 2 respectively:
Notation 1:
\begin{align}
y_{n_1}&=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\sum_{k=0}^{\frac{N}{2}-1}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}\tag{$\mathbf{1a}$}\\
&=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\sum_{k=0}^{\frac{N}{2}-1}Y_k^* \overbrace{e^{j2\pi n}}^{=1} e^{-j\frac{2\pi nk}{N}}\\
&=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\left(\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}\right)^*\\
&=y_{n/2}+y^*_{n/2}\tag{$\scriptstyle{\text{$y_{n/2}=y_n$ for $n=0, \ldots, N/2 -1$}}$}\\
&=2y_{\frac n2}\tag{$\scriptstyle{\text{if $y_n$ is real, then $y_{n/2}$ is real}}$}
\end{align}
Notation 2:
\begin{align}
y_{n_2}&=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\sum_{k=1}^{\frac{N}{2}}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}\tag{$\mathbf{1b}$}\\
&=\underbrace{\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+Y_N+\sum_{k=1}^{\frac{N}{2}-1}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}}_{\textrm{same as notation 1}}-Y_N-Y_{N/2}\\
&=2y_{n/2}-Y_N-Y_{N/2}\\
\end{align}
For Notation 3 however, the equation would be identical to equation $(1)$ if you changed the upper limit of the second summation to $N-1$ instead of $N$, otherwise you have an additional term as a result of the $N^{\rm th}$ term as shown below:
\begin{align}
y_{n_3}&=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\sum_{k=\frac{N}{2}}^{N}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}\tag{$\mathbf{1c}$}\\
&=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\left(\sum_{k=N/2}^{N-1}Y_k e^{j\frac{2\pi nk}{N}}\right)^*+Y_0\\
&=y_{n/2}+\left(y_{n/2}^{'}\right)^*+Y_0\tag{$\scriptstyle{\text{$y_{n/2}^{'}=y_n$ for $n=N/2, \ldots, N-1$}}$}\\
&=y_{n/2}+y_{n/2}^{'}+Y_0\\
&=y_n + Y_0
\end{align}
So, the correct notation should be:
$$
y_n=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\sum_{k=\frac{N}{2}}^{N-1}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}\tag{$\mathbf{2}$}
$$
UPDATE:
Regardless of equation $(1)$, to test Notation 1 I will maintain my last notation and go proving if the equality below holds, i.e. if equation $(\mathbf{2})$ equals equation $(\mathbf{1a})$ in Notation 1.
\begin{align}
\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\sum_{k=\frac{N}{2}}^{N-1}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}&=\sum_{k=0}^{\frac{N}{2}-1}Y_k e^{j\frac{2\pi nk}{N}}+\sum_{k=0}^{\frac{N}{2}-1}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}\\
\implies\sum_{k=\frac{N}{2}}^{N-1}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}&=\sum_{k=0}^{\frac{N}{2}-1}Y_{N-k} e^{j\frac{2\pi n(N-k)}{N}}\\
\end{align}
One way could be testing if the following equalities are satisfied:
\begin{align}
Y_{N/2}e^{j\frac{2\pi n(N/2)}{N}}&=Y_Ne^{j\frac{2\pi n(N)}{N}}\\
Y_{N/2-1}e^{j\frac{2\pi n(N/2-1)}{N}}&=Y_{N-1}e^{j\frac{2\pi n(N-1)}{N}}\\
Y_{N/2-2}e^{j\frac{2\pi n(N/2-2)}{N}}&=Y_{N-2}e^{j\frac{2\pi n(N-2)}{N}}\\
\vdots&=\vdots\\
Y_2e^{j\frac{4\pi n}{N}}&=Y_{N/2+2}e^{j\frac{2\pi n(N/2+2)}{N}}\\
Y_1e^{j\frac{2\pi n}{N}}&=Y_{N/2+1}e^{j\frac{2\pi n(N/2+1)}{N}}\\
\end{align}
It's neither the periodicity nor the Hermitian property of the DFT (for a real-valued signal i.e $Y_{N-k}=Y_{-k}=Y_k^*$) that we're seeing in the above equations. But let's check the first two and the last one for instance:
For the first equation:
\begin{align}
&Y_{N/2}e^{j\frac{2\pi n(N/2)}{N}}=e^{j\frac{2\pi n(N/2)}{N}}\left(\sum_{n=0}^{N-1}y[n]e^{-j\frac{2\pi(N/2)n}N}\right)=- \sum_{n=0}^{N-1}y[n]e^{-j\pi n}=\sum_{n=0}^{N-1}y[n]=Y_0\\
&Y_{N}e^{j\frac{2\pi n(N)}{N}}=e^{j\frac{2\pi n(N)}{N}}\left(\sum_{n=0}^{N-1}y[n]e^{-j\frac{2\pi(N)n}N}\right)=\sum_{n=0}^{N-1}y[n]e^{-j2\pi n}=\sum_{n=0}^{N-1}y[n]=Y_0\\
&\implies Y_{N/2}e^{j\frac{2\pi n(N/2)}{N}}= Y_{N}e^{j\frac{2\pi n(N)}{N}}\\
\end{align}
For the second equation:
\begin{align}
&Y_{N/2-1}e^{j\frac{2\pi n(N/2-1)}{N}}=e^{j\frac{2\pi n(N/2-1)}{N}}\left(\sum_{n=0}^{N-1}y[n]e^{-j\frac{2\pi(N/2-1)n}N}\right)= e^{-j\frac{2\pi n}{N}}\sum_{n=0}^{N-1}y[n]e^{j\frac{2\pi n}N}=e^{-j\frac{2\pi n}{N}}Y_1^*\\
&Y_{N-1}e^{j\frac{2\pi n(N-1)}{N}}=e^{j\frac{2\pi n(N-1)}{N}}\left(\sum_{n=0}^{N-1}y[n]e^{-j\frac{2\pi(N-1)n}N}\right)=e^{-j\frac{2\pi n}{N}} \sum_{n=0}^{N-1}y[n]e^{\frac{j2\pi n}N}=e^{-j\frac{2\pi n}{N}}Y_1^*\\
&\implies Y_{N/2-1}e^{j\frac{2\pi n(N/2-1)}{N}}=Y_{N-1}e^{j\frac{2\pi n(N-1)}{N}}\\
\end{align}
For the last equation:
\begin{align}
&Y_1e^{j\frac{2\pi n}{N}}=e^{j\frac{2\pi n}{N}}Y_1\\
&Y_{N/2+1}e^{j\frac{2\pi n(N/2+1)}{N}}=e^{j\frac{2\pi n(N/2+1)}{N}}\left(\sum_{n=0}^{N-1}y[n]e^{-j\frac{2\pi(N/2+1)n}N}\right)=e^{j\frac{2\pi n}{N}} \sum_{n=0}^{N-1}y[n]e^{\frac{-j2\pi n}N}=e^{j\frac{2\pi n}{N}}Y_1\\
&\implies Y_1e^{j\frac{2\pi n}{N}}=Y_{N/2+1}e^{j\frac{2\pi n(N/2+1)}{N}}
\end{align}
So, it looks like the initial notation in equation $(\mathbf{1a})$ is indeed identical to equation $(\mathbf{2})$. In conclusion, Notation 1 is correct for a real-valued signal. And Notation 3 need a slight change in the summation limit of the second term to be true. That's my two cents.
