Using the proper locale allows one to avoid changing decimal symbols:

DecimalFormat nf = (DecimalFormat) NumberFormat.getInstance(Locale.GERMANY);
nf.setParseBigDecimal(true);
BigDecimal bd = (BigDecimal) nf.parse("2,6");

And this ensures the scale of 2 is applied:

bd = bd.setScale(2, RoundingMode.HALF_UP); //2.60

If formatting the number as string in the same locale is required, the following should work (and perhaps preferred to bd.toString()):

NumberFormat nf2 = NumberFormat.getNumberInstance(Locale.GERMANY);
nf2.setMinimumFractionDigits(2);
//outputs "2,60"

I used DecimalFormat for formatting the BigDecimal instead of formatting the String, seems no problems with it.

The code is something like this:

bd = bd.setScale(2, BigDecimal.ROUND_DOWN);

DecimalFormat df = new DecimalFormat();
            
df.setMaximumFractionDigits(2);
            
df.setMinimumFractionDigits(0);
            
df.setGroupingUsed(false);

String result = df.format(bd);

The BigDecimal(double) constructor can have unpredictable behaviors. It is preferable to use BigDecimal(String) or BigDecimal.valueOf(double).

System.out.println(new BigDecimal(135.69)); //135.68999999999999772626324556767940521240234375
System.out.println(new BigDecimal("135.69")); // 135.69
System.out.println(new BigDecimal("13۵.۶9")); // 135.69
System.out.println(BigDecimal.valueOf(135.69)); // 135.69

The documentation for BigDecimal(double) explains in detail:

1. The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.
2. The String constructor, on the other hand, is perfectly predictable: writing new BigDecimal("0.1") creates a BigDecimal which is exactly equal to 0.1, as one would expect. Therefore, it is generally recommended that the String constructor be used in preference to this one.
3. When a double must be used as a source for a BigDecimal, note that this constructor provides an exact conversion; it does not give the same result as converting the double to a String using the Double.toString(double) method and then using the BigDecimal(String) constructor. To get that result, use the static valueOf(double) method.

This line is your problem:

litersOfPetrol = Float.parseFloat(df.format(litersOfPetrol));

There you formatted your float to string as you wanted, but but then that string got transformed again to a float, and then what you printed in stdout was your float that got a standard formatting. Take a look at this code

import java.text.DecimalFormat;

String stringLitersOfPetrol = "123.00";
System.out.println("string liters of petrol putting in preferences is "+stringLitersOfPetrol);
Float litersOfPetrol=Float.parseFloat(stringLitersOfPetrol);
DecimalFormat df = new DecimalFormat("0.00");
df.setMaximumFractionDigits(2);
stringLitersOfPetrol = df.format(litersOfPetrol);
System.out.println("liters of petrol before putting in editor : "+stringLitersOfPetrol);

And by the way, when you want to use decimals, forget the existence of double and float as others suggested and just use BigDecimal object, it will save you a lot of headache.

I think that the RoundingMode you are looking for is HALF_EVEN. From the javadoc:

Rounding mode to round towards the "nearest neighbor" unless both neighbors are equidistant, in which case, round towards the even neighbor. Behaves as for ROUND_HALF_UP if the digit to the left of the discarded fraction is odd; behaves as for ROUND_HALF_DOWN if it's even. Note that this is the rounding mode that minimizes cumulative error when applied repeatedly over a sequence of calculations.

Here is a quick test case:

BigDecimal a = new BigDecimal("10.12345");
BigDecimal b = new BigDecimal("10.12556");

a = a.setScale(2, RoundingMode.HALF_EVEN);
b = b.setScale(2, RoundingMode.HALF_EVEN);

System.out.println(a);
System.out.println(b);

Correctly prints:

10.12
10.13

UPDATE:

setScale(int, int) has not been recommended since Java 1.5, when enums were first introduced, and was finally deprecated in Java 9. You should now use setScale(int, RoundingMode) e.g:

setScale(2, RoundingMode.HALF_EVEN)

value = value.setScale(2, RoundingMode.CEILING)

The reason of such behaviour is that the string that is printed is the exact value - probably not what you expected, but that's the real value stored in memory - it's just a limitation of floating point representation.

According to javadoc, BigDecimal(double val) constructor behaviour can be unexpected if you don't take into consideration this limitation:

The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.

So in your case, instead of using

double val = 77.48;
new BigDecimal(val);

use

BigDecimal.valueOf(val);

Value that is returned by BigDecimal.valueOf is equal to that resulting from invocation of Double.toString(double).