Variable does not increase and I do not see the error

Once again you have a logic problem in your algorithm...

You're looking for matches, and you're doing it wrong...

Let's analyze your code (again, we've already done this several times):

for (int i = 0; i < matriz.length; i++) {
//recorres las filas de la matriz
    for (int j = 0; j < matriz[i].length; j++) {
    //recorres las columnas de la matriz
        if (vector.length  <= matriz[i].length ) {
        //compruebas si el vector es menor que la fila de la matriz, bien
            for (int k = 0; k < vector.length; k++) {
            //recorres el vector....
                if (vector[k] == matriz[i][j]) {
                //comparas el vector, contra la posicion de la matriz.. pero todo el vector?
                //contra la misma posicion de la matriz? entonces aqui esta el error.

Your solution should be something like this:

int[][] matriz = {{2, 4, 6}, {1, 2, 4, 5, 5}, {1, 8, 9}};
int[] vector = {2, 4, 5};

int punteroalvector = 0;
for (int i = 0; i < matriz.length; i++) 
{
   for (int j = 0; j < matriz[i].length; j++) 
   {
       if (vector[punteroalvector] == matriz[i][j]) 
       {
           punteroalvector++;
           if (punteroalvector == vector.length)
           {
              System.out.println("encontre");
              System.out.println("i: " + i + " j desde: " + (j-punteroalvector) + " j hasta: " + j);
              break;
            }
        }
        else
        {
            punteroalvector = 0;
        }
    }
    if (punteroalvector == vector.length)
    {

         break;
    }
}     

First, the problem of why it's not working for you:

If you look, you have 3 fornested, and where you make the comparison is in the innermost one, that is, you are comparing each of the elements of the vector with the first of the elements of the row of the matrix. Then all the elements of the vector with the second in the row of the matrix, and so on. The problem is that there is an iterator left over. You must iterate through the i-th row comparing the j-th elements of the vector and of that row.

I think this code should solve the problem.

int[][] matriz = {{2, 4, 6}, {1, 2, 4, 5, 5}, {1, 8, 9}};
int[] vector = {2, 4, 5};

int coincidencia = 0;

for (int i = 0; i < matriz.length; i++) {

    if (vector.length  == matriz[i].length ) {
        //si en esta fila no hay la misma cantidad de elementos, pasa a la siguiente

        coincidencia = 0;   //reset de la variable que actua como contador
        for (int j = 0; j < matriz.length; j++) {

            if (vector[j] == matriz[i][j]) {
                System.out.println("Coincidencia en " + "[" + i + "]" + "[" + j + "]");
                coincidencia++;
                System.out.println(coincidencia);
            } else {
                //si uno fallo, ya no es el vector lo que hay en esta fila
                break;
            }

        }

        if (coincidencia == vector.length) {
            //encontramos el vector!
            break;
        }   
    }   
}

The other problem I see in the code you're using is that you have the ifnested inside the second for, so it makes row size comparisons that are unnecessary.

And finally, the condition of the ifactually is that the row of the matrix is ​​exactly equal to the size of the vector you are looking for. This will speed up and simplify the code a lot.

I hope I have helped.

Excuse me, by coincidence you mean that the right vector is found in the matrix? If that is the case, it is much easier this way:

int[][] matriz = {{2, 4, 6}, {1, 2, 4, 5, 5}, {1, 8, 9}};
    int[] vector = {2, 4, 6};

    int coincidencias = 0;
    for (int i = 0; i < matriz.length; i++) {
        if(Arrays.equals(matriz[i], vector)){//Claramente Arrays.equals compara 2 arreglos de cualquier tipo para ver si sus elementos son los mismos.
            coincidencias++;
        }
    }

    JOptionPane.showMessageDialog(null, "Coincidencias: "+coincidencias);