$\newcommand\R{\mathbb R}\newcommand\1{\mathbf1}$When you say "I want to integrate over $\mathbb{R}^k$ restricted to the plain where $\sum_{i=1}^{k}y_i=x$", you have to specify the measure over the plane over which you want to integrate. 

It appears you want this measure to be induced by the Lebesgue measure on $\R^k$. Then the integration can be done as follows. Let 
$$c:=N^2r\in(0,\infty)$$
and 
$$t:=x/\sqrt k,$$
the (signed) distance from your plane 
$$\Pi_t:=\{y\in\R^k\colon u\cdot y=t\}=\{y\in\R^k\colon \1\cdot y=x\},$$
where $\cdot$ denotes the dot product, $\1:=(1,\dots,1)\in\R^k$, and 
$$u:=\1/\sqrt k$$
is a unit normal vector to the plane $\Pi_t$. Thus, instead of the parameter $x$, we use the more geometrical parameter $t$. 

Then the integral in question can be written as 
$$I_t:=e^{ct^2}J_t,\quad\text{where}\quad J_t:=\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2},$$
$|y|$ is the Euclidean norm of $y$, and $\mu_t$ is an appropriate measure over the plane $\Pi_t$. Specifically, the measures $\mu_t$, induced by the Lebesgue measure on $\R^k$, should be such that for all real $a$ and $b$ such that $a<b$ we have 
$$\int_a^b dt\, J_t=\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)e^{-c|y|^2}
=K_{a,b}:=\int_{\Pi_{a,b}}dy\,e^{-c|y|^2},$$
where 
$$\Pi_{a,b}:=\bigcup_{t\in[a,b]}\Pi_t=\{y\in\R^k\colon a\le u\cdot y\le b\}.$$

To compute the integral $K_{a,b}$, let us use a substitution of the form $y=Qz$, where $Q$ is any orthogonal $k\times k$ matrix whose first column is the unit vector $u$, so that 
$y=Qz$ implies $z_1=u\cdot y$, where $z_j$ is the $j$'s coordinate of the vector $z$; such an orthogonal matrix $Q$ exists. Then we can write 
$$K_{a,b}=\int_{\R^k}dy\,e^{-c|y|^2}\,1(a\le u\cdot y\le b) \\
=\int_{\R^k}dz\,e^{-c|z|^2}\,1(a\le z_1\le b) \\
=\int_a^b dz_1\,e^{-cz_1^2}\int_{\R^{k-1}}dw\,e^{-c|w|^2} \\
=\int_a^b dz_1\,e^{-cz_1^2}\,(\pi/c)^{(k-1)/2}.
$$
So, 
$$J_t=\frac d{dt}\,K_{a,t}=e^{-ct^2}\,(\pi/c)^{(k-1)/2}.$$
Thus, the integral in question is 
$$I_t:=e^{ct^2}J_t=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2}.$$