One way to get the SoP form starts by multiplying everything out, using the distributive law:
$$\begin{align*}
(ac+b)(a+b'c)+ac&=ac(a+b'c)+b(a+b'c)+ac\\
&=aca+acb'c+ba+bb'c+ac\\
&=ac+ab'c+ab+ac\\
&=ac+ab'c+ab\;.
\end{align*}$$
Then make sure that every term contains each of $a,b$, and $c$ by using the fact that $x+x'=1$:
$$\begin{align*}
ac+ab'c+ab&=ac(b+b')+ab'c+ab(c+c')\\
&=abc+ab'c+ab'c+abc+abc'\\
&=abc+ab'c+abc'\;.
\end{align*}$$
Alternatively, you can make what amounts to a truth table for the expression:
$$\begin{array}{cc}
a&b&c&ac+b&b'c&a+b'c&ac&(ac+b)(a+b'c)+ac\\ \hline
0&0&0&0&0&0&0&0\\
0&0&1&0&1&1&0&0\\
0&1&0&1&0&0&0&0\\
0&1&1&1&0&0&0&0\\
1&0&0&0&0&1&0&0\\
1&0&1&1&0&1&1&1\\
1&1&0&1&1&1&0&1\\
1&1&1&1&0&1&1&1
\end{array}$$
Now find the rows in which the expression evaluates to $1$; here it’s the last three rows. For a product for each of those rows; if $x$ is one of the variables, use $x$ if it appears with a $1$ in that row, and use $x'$ if it appears with a $0$. Thus, the last three rows yield (in order from top to bottom) the terms $ab'c$, $abc'$ and $abc$.
You can use the truth table to get the PoS as well. This time you’ll use the rows in which the expression evaluates to $0$ — in this case the first five rows. Each row will give you a factor $x+y+z$, where $x$ is either $a$ or $a'$, $y$ is either $b$ or $b'$, and $z$ is either $c$ or $c'$. This time we use the variable if it appears in that row with a $0$, and we use its negation if it appears with a $1$. Thus, the first row produces the sum $a+b+c$, the second produces the sum $a+b+c'$, and altogether we get
$$(a+b+c)(a+b+c')(a+b'+c)(a+b'+c')(a'+b+c)\;.\tag{1}$$
An equivalent procedure that does not use the truth table is to begin by using De Morgan’s laws to negate (invert) the original expression:
$$\begin{align*}
\Big((ac+b)(a+b'c)+ac\Big)'&=\Big((ac+b)(a+b'c)\Big)'(ac)'\\
&=\Big((ac+b)'+(a+b'c)'\Big)(a'+c')\\
&=\Big((ac)'b'+a'(b'c)'\Big)(a'+c')\\
&=\Big((a'+c')b'+a'(b+c')\Big)(a'+c')\\
&=(a'b'+b'c'+a'b+a'c')(a'+c')\\
&=a'b'(a'+c')+b'c'(a'+c')+a'b(a'+c')+a'c'(a'+c')\\
&=a'b'+a'b'c'+a'b'c'+b'c'+a'b+a'bc'+a'c'+a'c'\\
&=a'b'+a'b'c'+b'c'+a'b+a'bc'+a'c+a'c'\\
&=a'b'+b'c'+a'b+a'(c+c')\\
&=a'b+b'c'+a'b+a'\\
&=b'c'+a'\;,
\end{align*}$$
where in the last few steps I used the absorption law $x+xy=x$ a few times. Now find the SoP form of this:
$$\begin{align*}
b'c'+a'&=b'c'(a+a')+a'(b+b')(c+c')\\
&=ab'c'+a'b'c'+a'b(c+c')+a'b'(c+c')\\
&=ab'c'+a'b'c'+a'bc+a'bc'+a'b'c+a'b'c'\\
&=ab'c'+a'b'c'+a'bc+a'bc'+a'b'c\;.
\end{align*}$$
Now negate (invert) this last expression, and you’ll have the PoS form of the original expression:
$$\begin{align*}
(ab'c'&+a'b'c'+a'bc+a'bc'+a'b'c)'\\
&=(ab'c')'(a'b'c')'(a'bc)'(a'bc')'(a'b'c)'\\
&=(a'+b+c)(a+b+c)(a+b'+c')(a+b'+c)(a+b+c')\;,
\end{align*}$$
which is of course the same as $(1)$, though the factors appear in a different order.
The following three-valued $\{0, 1, X\}$ disjunctive truth table combines two inputs to one output:
f g | f+g
----+----
0 0 | 0
0 1 | 1
0 X | X
1 0 | 1
1 1 | 1
1 X | 1
X 0 | X
X 1 | 1
X X | X
Whenever at least one input is $true$, the output becomes $true$. Both inputs have to be $false$ to make the output $false$. Input combination of $false$ and don't care leads to output don't care.
Applied to your problem:
Resulting minimized expression for $f + g$:
x3' + x1' x2' + x1 x2 x5 + x2 x4 x5'
Best Answer
$$ab'C_{in}'+ a'b'C_{in} + abC_{in} + a'bC_{in}'$$ $$C_{in}'(ab'+a'b) + C_{in}(a'b'+ab)$$ This is of the form: $$X'Y + XY'$$ Which is: $$X \oplus Y$$ where, $$X = C_{in}$$ $$Y = (ab'+a'b) = (a \oplus b)$$
So finally you arrive at: $$ a \oplus b \oplus C_{in}$$
Hope the answer is clear !!