char digit = (char)(tmp.charAt(i) - '0');

In the ascii table, characters from '0' to '9' are contiguous. So, if you know that tmp.charAt(i) will return a character between 0 and 9, then subracting 0 will return the offset from zero, that is, the digit that that character represents.

All in all, that's a pretty good implementation in a number of key details.

Using the Character.isDigit() and Character.getNumericValue() methods are good to see.

The Math.* methods that handle the overflow conditions are good too.

I am not sure if you intended it, but you also correctly handle an obscure edge-case in 32-bit signed integer systems (not just Java), where Integer.MIN_VALUE is not the same as - Integer.MAX_VALUE ... and your code actually gets it right for an exact input of the text "-2147483648"

So, you have good details in your code.... and I can't see any broken edge-cases.

My only recommendation would be that a state-machine may make things simpler... with just one loop..... but the state-machine may be a bit messy too, although I think it works out better in the long run...

public static int rlAtoi(String str) {
    boolean started = false;
    boolean negative = false;
    int result = 0;
    try {
        for (char c : str.toCharArray()) {
            if (!started && Character.isWhitespace(c)) {
                // great, ignore it.
            } else if (!started && (c == '+' || c == '-')) {
                // great, a sign
                negative = c == '-';
                started = true;
            } else if (Character.isDigit(c)) {
                result = Math.multiplyExact(result, 10);
                result = Math.addExact(result, Character.getNumericValue(c));
                started = true;
            } else {
                // done....
                break;
            }
        }
    } catch (ArithmeticException e) {
        return negative ? Integer.MIN_VALUE : Integer.MAX_VALUE;
    }
    return negative ? -result : result;
}

Note that in a raw performance benchmark, I suspect your solution will be (slightly) faster, but I prefer readability over small incremental performance gains, unless performance is extremely critical.

Normally when parsing a String into an int, people expect it to be a static method

Having to create a class to do the computation is a bit bulky, and then you need to worry about synchronization (what happens if this class is shared across multiple threads?)

boolean isOverflow(int prev, int curr) {
    // prev = value at last iteration
    // curr = value after current iteration
    if (curr/10 == prev) {
        return false;
    }
    return true;
}

This method is already a great candidate to become static, it only relies on its two inputs to figure out if an overflow happened

boolean isNegative = false;
int i = 0; // string iterator position

void checkSign(String str) //uses i and isNegative

int signSolution(int solution) //uses isNegative

These are slightly more tricky, int signSolution(int solution) can just take another parameter: static int signSolution(int solution, boolean isNegative)

void checkSign(String str) would need to be included as part of the myAtoI method or be rewritten a bit

If we treat + as whitespace, then it would only need to worry about handling -

static boolean checkSign(String str, int i)
{
    return str.charAt(i) == '-';
}

And calls to checkSign(str); would instead be if(checkSign(str)) i++;

Be consistent. You're mixing for and while loops of similar structure

while (i < str.length() && str.charAt(i) == ' ')
    i++;
[...]
for (; i < str.length(); i++) {
    [...]
}

They can either both be for loops, or both be while loops, but mixing is unexpected

for(; i < str.length() && str.charAt(i) == ' '; i++);
[...]
for(; i < str.length(); i++)
{
    [...]
}

vs

while(i < str.length() && str.charAt(i) == ' ')
    i++;
[...]
while(i < str.length())
{
    [...]
    i++;
}

Checking "whitespace" currently only checks spaces

Instead of using str.charAt(i) == ' ', you can offload complicated functionality to another method (and be easier to change later)

static boolean isWhiteSpace(char c)
{
    return c == ' ';
}

Then if you'd like to include all whitespace (and + and mentioned above)

static boolean isWhiteSpace(char c)
{
    return Character.isWhitespace(c) || (c == '+');
}

Instances of if(bool) return true; else return false; are the same as just return bool;

if (curr/10 == prev) {
        return false;
}
return true;

Can be simplified to just

return curr / 10 != prev;

OptionalA:

Another way to check for overflow is to store the output as a long, and check if it overflows

public static int myAtoI(String str)
{
    long solution = 0;
    [...]
}

private static isOverflow(long num)
{
    return num == (int)num;
}

OptionalB:

Instead of adding the values and then flipping the result, you could subtract values if it's a negative number