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The desired integral $I(x)$ can be written as $^\ast$

$$I(x)=e^{N^2rx^2/k}J(x),\;\;J(x)=\mathop{\idotsint}\delta\left(x-{\textstyle{\sum_{j=1}^{k}}y_j}\right) e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}.$$ Fourier transform $J(x)$, $$F(q)=\int_{-\infty}^\infty e^{iqx}J(x)\,dx=\mathop{\idotsint} e^{-N^2r\sum_{j=1}^{k}y_j^2}e^{\sum_{j=1}^k qy_j} dy_1\dots dy_{k}=N^{-k}(\pi/r)^{k/2}e^{-kq^2/(4N^2r)}.$$ Now Fourier transform back from $F(q)$ to $J(x)$ and you're done: $$J(x)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-iqx}F(q)\,dq=k^{-1/2} (N^2 r/\pi)^{(1-k)/2} e^{-N^2 r x^2/k}$$ $$\Rightarrow I(x)=k^{-1/2} (N^2 r/\pi)^{(1-k)/2}.$$ This differs from both of the answers in the OP, but answer number 1 will agree if a typo is corrected ($\pi^k$ should be $\pi^{k/2}$, which is the factor coming from the Gaussian integrals in method 1).

$^\ast$ As noted by Iosif Pinelis, the notation $\idotsint_{\sum_{i=1}^{k}y_i=x}$ in the OP is ambiguous, I replaced it by the delta function $\idotsint \,\delta(x-\sum_{i=1}^{k}y_i)$ to remove the ambiguity.
More generally, one can define $I_a(x)$ and $J_a(x)$ with the delta function $\delta(ax-a\sum_{i=1}^{k}y_i)$ for any $a>0$. This amounts to the same Euclidian measure on the hyperplane, but the distance from the hyperplane is rescaled by a factor $a$. Since $\delta(au)=a^{-1}\delta(u)$, one has $$I_a(x)=a^{-1}I_1(x)=a^{-1}k^{-1/2}(N^2r/\pi)^{(1-k)/2}.$$ Iosif's answer corresponds to the choice $a=k^{-1/2}$.

The desired integral $I$ can be written as $^\ast$

$$I=e^{N^2rx^2/k}J(x),\;\;J(x)=\mathop{\idotsint}\delta\left(x-{\textstyle{\sum_{j=1}^{k}}y_j}\right) e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}.$$ Fourier transform $J(x)$, $$F(q)=\int_{-\infty}^\infty e^{iqx}J(x)\,dx=\mathop{\idotsint} e^{-N^2r\sum_{j=1}^{k}y_j^2}e^{\sum_{j=1}^k qy_j} dy_1\dots dy_{k}=N^{-k}(\pi/r)^{k/2}e^{-kq^2/(4N^2r)}.$$ Now Fourier transform back from $F(q)$ to $J(x)$ and you're done: $$J(x)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-iqx}F(q)\,dq=k^{-1/2} (N^2 r/\pi)^{(1-k)/2} e^{-N^2 r x^2/k}$$ $$\Rightarrow I=k^{-1/2} (N^2 r/\pi)^{(1-k)/2}.$$ This differs from both of the answers in the OP, but answer number 1 will agree if a typo is corrected ($\pi^k$ should be $\pi^{k/2}$, which is the factor coming from the Gaussian integrals in method 1).

$^\ast$ As noted by Iosif Pinelis, the notation $\idotsint_{\sum_{i=1}^{k}y_i=x}$ in the OP is ambiguous, I replaced it by the delta function $\idotsint \,\delta(x-\sum_{i=1}^{k}y_i)$ to remove the ambiguity.
More generally, one can define $I_a$ and $J_a(x)$ with the delta function $\delta(ax-a\sum_{i=1}^{k}y_i)$ for any $a>0$. This amounts to the same Euclidian measure on the hyperplane, but the distance from the hyperplane is rescaled by a factor $a$. Since $\delta(au)=a^{-1}\delta(u)$, one has $$I_a=a^{-1}I_1(x)=a^{-1}k^{-1/2}(N^2r/\pi)^{(1-k)/2}.$$ Iosif's answer corresponds to the choice $a=k^{-1/2}$.

The desired integral $I(x)$ can be written as $^\ast$

$$I(x)=e^{N^2rx^2/k}J(x),\;\;J(x)=\mathop{\idotsint}\delta\left(x-{\textstyle{\sum_{j=1}^{k}}y_j}\right) e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}.$$ Fourier transform $J(x)$, $$F(q)=\int_{-\infty}^\infty e^{iqx}J(x)\,dx=\mathop{\idotsint} e^{-N^2r\sum_{j=1}^{k}y_j^2}e^{\sum_{j=1}^k qy_j} dy_1\dots dy_{k}=N^{-k}(\pi/r)^{k/2}e^{-kq^2/(4N^2r)}.$$ Now Fourier transform back from $F(q)$ to $J(x)$ and you're done: $$J(x)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-iqx}F(q)\,dq=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2} e^{-N^2 r x^2/k}$$ $$\Rightarrow I(x)=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2}.$$ This differs from both of the answers in the OP, but answer number 1 will agree if a typo is corrected ($\pi^k$ should be $\pi^{k/2}$, which is the factor coming from the Gaussian integrals in method 1).

The desired integral $I(x)$ can be written as $^\ast$

$$I(x)=e^{N^2rx^2/k}J(x),\;\;J(x)=\mathop{\idotsint}\delta\left(x-{\textstyle{\sum_{j=1}^{k}}y_j}\right) e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}.$$ Fourier transform $J(x)$, $$F(q)=\int_{-\infty}^\infty e^{iqx}J(x)\,dx=\mathop{\idotsint} e^{-N^2r\sum_{j=1}^{k}y_j^2}e^{\sum_{j=1}^k qy_j} dy_1\dots dy_{k}=N^{-k}(\pi/r)^{k/2}e^{-kq^2/(4N^2r)}.$$ Now Fourier transform back from $F(q)$ to $J(x)$ and you're done: $$J(x)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-iqx}F(q)\,dq=k^{-1/2} (N^2 r/\pi)^{(1-k)/2} e^{-N^2 r x^2/k}$$ $$\Rightarrow I(x)=k^{-1/2} (N^2 r/\pi)^{(1-k)/2}.$$ This differs from both of the answers in the OP, but answer number 1 will agree if a typo is corrected ($\pi^k$ should be $\pi^{k/2}$, which is the factor coming from the Gaussian integrals in method 1).

$^\ast$ As noted by Iosif Pinelis, the notation $\idotsint_{\sum_{i=1}^{k}y_i=x}$ in the OP is ambiguous, I replaced it by the delta function $\idotsint \,\delta(x-\sum_{i=1}^{k}y_i)$ to remove the ambiguity.
More generally, one can define $I_a(x)$ and $J_a(x)$ with the delta function $\delta(ax-a\sum_{i=1}^{k}y_i)$ for any $a>0$. This amounts to the same Euclidian measure on the hyperplane, but the distance from the hyperplane is rescaled by a factor $a$. Since $\delta(au)=a^{-1}\delta(u)$, one has $$I_a(x)=a^{-1}I_1(x)=a^{-1}k^{-1/2}(N^2r/\pi)^{(1-k)/2}.$$ Iosif's answer corresponds to the choice $a=k^{-1/2}$.

The desired integral $I(x)$ can be written as  $$I(x)=e^{N^2rx^2/k}J(x),\;\;J(x)=\mathop{\idotsint}\delta\left(x-{\textstyle{\sum_{j=1}^{k}}y_j}\right) e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}.$$ Fourier transform $J(x)$, $$F(q)=\int_{-\infty}^\infty e^{iqx}J(x)\,dx=\mathop{\idotsint} e^{-N^2r\sum_{j=1}^{k}y_j^2}e^{\sum_{j=1}^k qy_j} dy_1\dots dy_{k}=N^{-k}(\pi/r)^{k/2}e^{-kq^2/(4N^2r)}.$$ Now Fourier transform back from $F(q)$ to $J(x)$ and you're done: $$J(x)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-iqx}F(q)\,dq=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2} e^{-N^2 r x^2/k}$$ $$\Rightarrow I(x)=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2}.$$ This differs from both of the answers in the OP, but answer number 1 will agree if a typo is corrected ($\pi^k$ should be $\pi^{k/2}$, which is the factor coming from the Gaussian integrals in method 1).

The desired integral $I(x)$ can be written as $^\ast$

$$I(x)=e^{N^2rx^2/k}J(x),\;\;J(x)=\mathop{\idotsint}\delta\left(x-{\textstyle{\sum_{j=1}^{k}}y_j}\right) e^{-N^2r\sum_{j=1}^{k}y_j^2} dy_1\dots dy_{k}.$$ Fourier transform $J(x)$, $$F(q)=\int_{-\infty}^\infty e^{iqx}J(x)\,dx=\mathop{\idotsint} e^{-N^2r\sum_{j=1}^{k}y_j^2}e^{\sum_{j=1}^k qy_j} dy_1\dots dy_{k}=N^{-k}(\pi/r)^{k/2}e^{-kq^2/(4N^2r)}.$$ Now Fourier transform back from $F(q)$ to $J(x)$ and you're done: $$J(x)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-iqx}F(q)\,dq=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2} e^{-N^2 r x^2/k}$$ $$\Rightarrow I(x)=k^{-1/2} N^{1-k} (r/\pi)^{(1-k)/2}.$$ This differs from both of the answers in the OP, but answer number 1 will agree if a typo is corrected ($\pi^k$ should be $\pi^{k/2}$, which is the factor coming from the Gaussian integrals in method 1).