The result is false in general even for $d=1$. E.g., let
$$K=f_r+f_s,$$
where $f_t$ is the density of $N(0,t^2)$. Then for $x_i=y_i=i$ ($\forall i=1,\dots,n$) and
$$(n,r,s,x_*)=\Big(3,\frac{427}{215},\frac{1}{1547},\frac{472}{473}\Big)$$
we have
$$\hat y'(x_*)=-527.1\ldots<0.$$
Here is the graph $\{(x,\hat y(x))\colon\frac{471}{473}\le x\le\frac{475}{473}\}$:
We see a very narrow dip.
However, $\hat y'\ge0$ if $K$ is log concave. Indeed, letting
$$k_i:=K(x-x_i)\quad\text{and}\quad k'_i:=K'(x-x_i),$$
we have
$$
\begin{aligned}
2\Big(\sum_{i=1}^n k_i\Big)^2\hat y'(x)
&=\sum_{i,j=1}^n(k'_i y_i k_j-k_i y_i k'_j+k'_j y_j k_i-k_j y_j k'_i) \\
&=\sum_{i,j=1}^n(y_i-y_j)\Big(\frac{k'_i}{k_i}-\frac{k'_j}{k_j}\Big)k_ik_j\ge0,
\end{aligned}\tag{1}
$$
because $y_i$ is increasing in $i$ and
$$\frac{k'_i}{k_i}=(\ln K(x-x_i))' \tag{2}$$
is increasing in $i$; the latter holds because $x_i$ is increasing in $i$ and $(\ln K)'$ is decreasing (since $K$ is log concave). (In particular, any normal density is log concave.)
In the case $d>1$, the desired monotonicity fails to hold in general even when $K$ is log concave. E.g., let
$$K(u_1,\dots,u_d):=\exp\{u_1u_2-u_1^2-\cdots-u_d^2\},$$
$n=2$, $x_1=(0,\dots,0)$, $x_2=(0,1,0,\dots,0)$, $y_1=0$, and $y_2=1$.
Then $(\partial_1\ln K)(u_1,\dots,u_d)=u_2-2u_1$, where $\partial_1$ denotes the partial derivative with respect to the first coordinate. On the other hand (cf. (1) and (2)), $(\partial_1\hat y)(0,\dots,0)$ equals $l'_2-l'_1$ in sign, where
$$l'_1:=(\partial_1\ln K)(0,\dots,0)=0$$
and
$$l'_2:=(\partial_1\ln K)(0,-1,0,\dots,0)=-1<0=l'_1.$$
So, $(\partial_1\hat y)(0,\dots,0)<0$, as claimed.
Looking back at (1), it is clear that the necessary and sufficient condition for the desired monotonicity is that the appropriate analogs of $\frac{k'_i}{k_i}-\frac{k'_j}{k_j}$ be $\ge0$ whenever for the corresponding $x_i$ and $x_j$ we have $x_i\ge x_j$. In particular, this will happen if $K$ is the product of log-concave functions of one coordinate each.
Here we of course need to assume that that $x_i$'s are linearly ordered, so that, without loss of generality, $x_i\le x_j$ if $i<j$. Indeed, suppose e.g. that $x_1=(1,0,\dots,0)$, $y_1=1$, $x_i=(0,1,0,\dots,0)$ and $y_i=0$ for $i=2,\dots,n$. Then the implication $x_i\le x_j\implies y_i\le y_j$ holds for all $i,j$. This implication will also hold if we replace here $y_1$ by $-1$. But then $\hat y$ will change in sign, and hence its monotonicity pattern will change to the opposite one.