An image uploaded from html and processed with js/php/mysql does not display well

There are several errors in your code:

1. To send images you must declare the form with the attribute enctype="multipart/form-data". This literally means that the form has several parts , on the one hand the normal data such as name, surname, etc, which will be in the super global of the form method ( $_POST, $_GET...) and on the other hand the files, which will be in the super overall $_FILES.
2. Attachments are not as easily serialized by hand as other data. To simplify this process, the simplest thing is to use the FormData API , in the section Sending files using FormData objects you will see how easy it is to send multi-part forms to the server from Javascript.
3. To be able to use FormData without too much hassle, it's a good idea to give an attribute nameto every element you want to put information on.
4. On the server, something like this $datosJSON = $_POST["datos"];would not collect all the information traveling in a multi-part form. As already said in (1) the files are located in the super global $_FILES.
5. If the Ajax request expects a JSON in response, you cannot in any case issue this: die(mysqli_error($conexion), in case of error there will not be a JSON, but a string, you must control that by adding an error key in $respuesta.
6. This mysqli_query($conexion,"utf8")doesn't make any sense in your code. The charset you set with mysqli_set_charset().

We can apply what was said in your code as follows:

HTML

We give an attribute nameto the inputs, so FormData will automatically serialize all inputs that have data written to them without having to do it by hand. For simplicity we will change the attributes idto name.

<form name="frmAltaNoticia" id="frmAltaNoticia" method="POST" enctype="multipart/form-data">
            <div class="col-xs-3 col-sm-3 col-md-3">
                <label>IdNoticia: </label>
                <input type="text" placeholder="A0000"class="form-control" name="txtIdNoticia">
            </div>
            
            <div class="col-xs-3 col-sm-3 col-md-3">
                <label>Fecha: </label>
                <input type="date" placeholder="Fecha" class="form-control" name="txtFecha">
            </div>          
            
            <div class="col-xs-6 col-sm-6 col-md-6">
                <label>Título: </label>
                <input type="text" placeholder="Título"class="form-control" name="txtTitulo">
            </div>                                      
            <div class="col-xs-3 col-sm-3 col-md-3">
                <label>Autor: </label>
                <input type="text" placeholder="Autor"class="form-control" name="txtAutor">
            </div>  
            <div class="col-xs-9 col-sm-9 col-md-9">
                <label>Descripción:</label>
                <textarea name="txtDescripcion" placeholder="Descripción" class="form-control" id="txtDescripcion"></textarea>
            </div>

            <div class="col-xs-5 col-sm-5 col-md-5">
                <label>Imagen: </label>
                <input type="file" class="form-control" name="txtImagen">
            </div>  
                    
            <div class="col-md-3">
            <input class="form-check-input" type="checkbox" value="" name="invalidCheck" required>
            <label class="form-check-label" for="invalidCheck">Aceptar los términos</label>
            <div class="invalid-feedback">Debes aceptar los términos y condiciones</div>
            </div>

            <div class="col-xs-3 col-sm-3 col-md-3">
            <input type="button" class="btn btn-dark btn-lg"  id="btnAceptarNoticia" onclick="ValidarNoticia()" value="Aceptar"></input>
            </div>  
        </form>

JavaScript

There are several modifications here. Instead of the shortcut $.post, we'll use $.ajax, which allows us to more clearly set other parameters that are needed by FormData. It also allows us to handle a block doneand another failalways recommended for errors. In fact, the parameters you passed respuestaAltaNoticia()were not correct. One thing is the errors of the response and another the errors of the request itself that must always be handled in fail.

function crearNoticia()  {
    /* Esta línea basta para recoger todo lo que haya en el formulario*/
    var mData = new FormData(document.getElementById("frmAltaNoticia"));
    var mAjax = $.ajax({
        url: "../ApartadosPHP/altanoticias.php",
        method: "POST",
        data: mData,
        processData: false,  
        contentType: false,
        dataType: "json"
    });

    mAjax.done(function( oDatos ) {
        respuestaAltaNoticia(oDatos);
    });

    mAjax.fail(function( jqXHR, textStatus ) {
        alert( "Request failed: " + textStatus );
    });
}

function respuestaAltaNoticia(oDatos) {

    if (oDatos.error) 
    {
        alert(oDatos.mensaje);
    } 
    else 
    {
        alert(oDatos.mensaje);
        frmAltaNoticia.reset();
    }
}

PHP

# Obtenemos los datos normales con el estilo que usabas
$oNoticia = json_decode($_POST);


// Creamos la conexión al servidor.
$conexion = mysqli_connect($servidor, $usuario, $password,$basedatos);
if (!$conexion) { 
    $respuesta["error"]=1;
    #Considera cambiarlo por un mensaje personalizado en producción
    $respuesta["mensaje"]=mysqli_error($conexion);
} else {
    mysqli_set_charset($conexion,"utf8");
    # Pero como dijimos, la imagen está en otra super global
    $mFile = file_get_contents($_FILES['txtImagen']["tmp_name"]);
    $mBlob=$conexion->real_escape_string($mFile);
    $sql = "INSERT INTO noticias (idnoticia, titulo, descripcion, autor, fecha, imagen) VALUES ('$oNoticia->txtIdNoticia','$oNoticia->txtTitulo','$oNoticia->txtDescripcion','$oNoticia->txtAutor','$oNoticia->txtFecha','$mBlob');";
    $resultado = mysqli_query($conexion,$sql);

    if ($resultado){
        $respuesta["error"] = 0;
        $respuesta["mensaje"] = "Alta realizada"; 
    } else {
        $respuesta["error"] = 1;
        $respuesta["mensaje"] = "Error en el proceso de alta: ".mysqli_error($conexion);
    }
}

echo json_encode($respuesta);

mysqli_close($conexion);

PS: Security Warning

The INSERT query you are implementing here is highly vulnerable to SQL Injection attacks . A malicious user could manipulate any of the values ​​and prepare code that causes irreparable damage to your system or data. Consider using prepared queries to neutralize that risk.