Here are a couple of examples, the first with almost full detail, the second with less.
First I’ll convert $156_{16}$ to base ten using repeated division in base sixteen. I’ll use $A,B,C,D,E$, and $F$ for the base sixteen digits corresponding to base ten $10,11,12,13,14$, and $15$. I’ll also use a subscript $s=16$ to indicate that a number is to be interpreted in base sixteen.
Divide $156_s$ by $A_s$. Do this just as you would in base ten: $A_s$ won’t go into $1_s$, but it will go into $15_s$. In fact $15_s=1\cdot 16+5=21$, and $A_s=10$, so it goes twice. The first digit of your quotient is $2_s$, so you need to subtract $2_s\cdot A_s$ from $15_s$.
$2_s\cdot A_s=2\cdot 10=20=1\cdot16+4=14_s$, and $15_s-14_s=1_s$, so after you bring down the $6_s$, you’re left dividing $A_s$ into $16_s$.
Similarly, $16_s=1\cdot 16+6=22$, so $A_s$ goes in twice. After you repeat the previous step (with suitable minor modifications) you have your full quotient $22_s$ and overall remainder $2_s$, as shown below.
$\begin{array}{} &&&2&2\\ &&\text{_}&\text{_}&\text{_}\\ A&)&1&5&6\\ &&1&4\\ &&-&-&-\\ &&&1&6\\ &&&1&4\\ &&&-&-\\ &&&&\color{red}2 \end{array}$
Now divide $22_s$ by $A_s$. $22_s=2\cdot 16+2=34$, so the integer part of the quotient is $3_s$:
$\begin{array}{} &&&3\\ &&\text{_}&\text{_}\\ A&)&2&2\\ &&1&E\\ &&-&-\\ &&&\color{red}4\\ \end{array}$
Finally, divide this last quotient, $3_s$, by $A_s$:
$\begin{array}{} &&0\\ &&\text{_}\\ A&)&3\\ &&0\\ &&-\\ &&\color{red}3\\ \end{array}$
Read off the red remainders in reverse order: $156_s=342$.
Here’s one a little more complicated, the conversion of $2BA_s$ to base three.
$\begin{array}{ccccc|cccc|cccc|cccc|ccc} &&&E&8&&&4&D&&&1&9&&&&8&&&\color{red}2\\ &&\text{_}&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}&\text{_}&&&\text{_}\\ 3&)&2&B&A&3&)&E&8&3&)&4&D&3&)&1&9&3&)&8\\ &&2&A&&&&C&&&&3&&&&1&8&&&6\\ &&-&-&-&&&-&-&&&-&-&&&-&-&&&-\\ &&&1&A&&&2&8&&&1&D&&&&\color{red}1&&&\color{red}2\\ &&&1&8&&&2&7&&&1&B\\ &&&&-&&&-&-&&&-&-\\ &&&&\color{red}2&&&&\color{red}1&&&&\color{red}2 \end{array}$
That last quotient of $2$ is less than the divisor, so the next division will have a $0$ quotient and remainder of $\color{red}2$, so I’ve skipped the step and colored the quotient instead. Reading the remainders in reverse order, we have $2BA_s=221212_t$ (where the subscript $t$ indicates base three).
Check: $2BA_s=2\cdot 256+11\cdot16+10=698\;,$ and $221212_t=2\cdot 243+2\cdot81+1\cdot27+2\cdot9+1\cdot3+2=698\;.$