$site = Site::create([
    "path" => $request->path,
    "site_link" => $request->link,
]);

不返回最后一个插入id。。。尝试

$site = new Site;
$site->path = $request->path;
$site->site_link = $request->link;
$site->save();

关于这一部分：

if ($request->features) {
    $features = explode(',', $request->features);

    foreach ($features as $feature) {
        $site->features()->save(SiteFeature::create(["feature" => $feature]));
    }
}

如果上面的代码不起作用，您可以这样做。。。

if ($request->features) {
    $features = explode(',', $request->features);

    foreach ($features as $feature) {
        $sf = new SiteFeature;
        $sf->feature = $feature;
        $sf->site_id = $site->id;
        $sf->save();
    }
}

这个 Model::create 方法实际上在数据库中创建了一条记录，因为您没有指定所需的 site_id ，这是失败，这就是为什么你看到这个错误。
看来你是想利用拉威尔的关系来拯救一个新的世界 SiteFeature 但是您正在将已经存在的对象（如果查询没有失败）传递给关系的 save 方法。
您需要传递 SiteFeature ，尚未保存到数据库的 save 方法：

$this->features()->save(new SiteFeature(['feature' => $feature]));

或者你可以用关系的 create 方法为避免必须同时传入示例，只需提供以下属性：

$this->features()->create(['feature' => $feature]);

试试这个
在您的模型中 features ```
public function features()
{
return $this->hasMany('App\SiteFeature','site_id','id');
}

你是模特吗 `SiteFeature` ```
public function siteFeatures()
{
  return $this->belongsTo('App\features', 'site_id', 'id');
}