Problem displaying inserted form

In your code I encountered two problems.

1. First one is you have not defined $current_user and $abcde variable. For using global $current_user you need to define it in the top of the function as global $current_user. Otherwise it will not get the $current_user variable. And also for getting the current user ID yo need to use $current_user->ID after declaring the global variable. And for $abcde variable please declare or store the value first in that variable.
2. And second one is you are checking logic with if (!$alibaba) which actually means if the $alibaba variable is false. Obviously if the DB entry is successful the $alibaba will return true. So the loop will not be executed.

So now your whole code will be like below-

function insert_form() {

    global $wpdb, $current_user;
    $table_name = $wpdb->prefix . "food_resarved";
    $abcde=""; // Later some where put a value in it. Otherwise it will not work.
    $alibaba = $wpdb->insert(
        $table_name,
        array(
            'id' => '',
            'user_id' => wp_get_current_user()->ID,
            'name' => wp_get_current_user()->user_firstname,
            'lastname' => wp_get_current_user()->user_lastname,
            'post_title' => $_POST['post_title'],
            'food_selected' => $_POST['food'],
            'guest_number' => $_POST['guest_food'],
            'email' => wp_get_current_user()->user_email
        ),
        array( '%d', '%s', '%s' , '%s' , '%s' , '%s' , '%s', '%s' )
    );

    if ( $alibaba ) {
        $mylink = $wpdb->get_results( "select * from wp_food_resarved where user_id like '".$current_user->ID."' AND post_title like '".$abcde."' ");
        foreach ($mylink as $posta){
            echo '<div class="ordered-food">';
            echo ' شما ';
            echo $posta->food_selected;
            echo ' سفارش داده‌اید ';
            echo '</div>';
        }
    }
    else {
        echo "Die";
    }
}

Hope that helps you.