How to number the alphabet and add your letters to get a result for each written word?

What your code is doing is simply printing what someone else types. You need to add summation logic and alphabet mapping. For this, you can make the following modifications:

1- Create a method that counts the sums:

private static int calcularSomaPalavra(String palavra, Map<Character, Integer> alfabeto) {
    int valorSoma = 0;
    for (char caractere : palavra.toCharArray()) {
        if (isCaractereEncontrado(alfabeto, caractere))
            valorSoma += getValorCaractere(alfabeto, caractere);
    }
    return valorSoma;
}

What was done was to create a method that given a word gets the sum of the characters of the word, simple as well. The big balcony here is to check if the character has a numerical mapped value and add it to the totalizer. This can be done in many ways, for simplicity you can use Map :

// Declaração do map de letras do alfabeto.
Map<Character, Integer> alfabeto = new HashMap<Character, Integer>();

//Mapeamento dos valores de cada letra do alfabeto
alfabeto.put('a', 1);
alfabeto.put('b', 2);
...
alfabeto.put('z', 26);

private static Integer getValorCaractere(Map<Character, Integer> alfabeto, char caractere) {
    return alfabeto.get(caractere);
}

private static boolean isCaractereEncontrado(Map<Character, Integer> alfabeto, char caractere) {
    return getValorCaractere(alfabeto, caractere) != null;
}

Using% w / o%, the numerical value check logic of the character is only the call of the Map method.

2- Print the sum of the word:

int valorSoma = calcularSomaPalavra(soma, alfabeto);
System.out.println(valorSoma);

Now you're actually printing the sum, not just what the user types. Putting it all together, we would have:

public static void main(String[] args) {
    Map<Character, Integer> alfabeto = new HashMap<Character, Integer>();
    Scanner on = new Scanner(System.in);
    System.out.println("Digite a palavra: ");

    String soma;

    soma = on.nextLine();

    alfabeto.put('a', 1);
    alfabeto.put('b', 2);
    ...
    alfabeto.put('z', 26);

    int valorSoma = calcularSomaPalavra(soma, alfabeto);
    System.out.println(valorSoma);

}

private static int calcularSomaPalavra(String palavra, Map<Character, Integer> alfabeto) {
    int valorSoma = 0;
    for (char caractere : palavra.toCharArray()) {
        if (isCaractereEncontrado(alfabeto, caractere))
            valorSoma += getValorCaractere(alfabeto, caractere);
    }
    return valorSoma;
}

private static Integer getValorCaractere(Map<Character, Integer> alfabeto, char caractere) {
    return alfabeto.get(caractere);
}

private static boolean isCaractereEncontrado(Map<Character, Integer> alfabeto, char caractere) {
    return getValorCaractere(alfabeto, caractere) != null;
}

One tip is to always get a problem and break it down into resolution steps, so you can solve a bigger problem by solving each step of it, which makes it easier to work out the overall solution.

It's quite simple using math:

import java.util.Scanner;

class Ideone {
    public static void main(String[] args) {
        Scanner on = new Scanner(System.in); 
        System.out.println("Digite a palavra: ");
        String texto = on.nextLine();
        int soma = 0;
        for (char caractere : texto.toCharArray()) { //varre cada caractere
            //"A" vale 97, então tira 96 e assim por diante
            soma += caractere > 96 && caractere < 123 ? caractere - 96 : 0;
        }
        System.out.println(soma);    
    }  
}

See running on ideone .

Other things can be done differently, but the question does not give clear requirements.

The important thing is that this solution uses the principle of KISS . Never complicate what can be simple. There are problems that are inherently complex and we can only manage them well. Complicated solutions can and should be avoided.