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I tried to multiply $1111_2$ with $1111_2$ but I came across a problem, namely adding $1+1+1+1$ and $1+1+1+1+1$. Here is my method:

I can't figure out how to add the $3rd$ and $4th$ column. I thought that $1+1+1+1 = 4 equiv 100$ but how can I implement it here?

From the end, first digit: $color{red}1$.

2nd digit: $1+1=color{blue}1color{red}0$.

3rd digit: $1+1+1+color{blue}1=10+10=color{blue}{10}color{red}0$.

4th digit: $1+1+1+1+color{blue}{10}=10+10+10=100+10=color{blue}{11}color{red}0$.

5th digit: $1+1+1+color{blue}{11}=10+100=color{blue}{11}color{red}0$.

6th digit: $1+1+color{blue}{11}=1+100=color{blue}{10}color{red}1$.

7th digit: $1+color{blue}{10}=color{red}{11}$.

Collecting backwards: $1111_2times1111_2=11100001$.

You carry over $color{blue}{2=10_2}$, and then continue addition. It is that simple.

Remember, even if we are adding single-digit numbers while moving down a column, there is no guarantee that the carry over will be a single digit number. For example, if you perform , say:
$$
29 \ + \ 09 \ +\ 09 \+ \ vdots \ + \ 09
$$

where the number of $09$s is $12$, then the sum of the first column will be $117$, so your sum will be:
$$
2^{11}9 \ + \ 09 \+ \09 \ + \ + \ vdots \ + \ 09 \ ----- \ 2^{11}7 = 137 \ -----
$$

In this case, for example, we would have :
$$
begin{matrix}
color{pink}{^10}&color{brown}{^{10}0}&color{green}{^{11}0}&color{orange}{^{11}0}&color{red}{^{10}1}&color{blue}{^11}&1&1 \+&&&&&&&\ 0&0&0&1&1&1&1&0\ +&&&&&&& \ 0&0&1&1&1&1&0&0 \ +&&&&&&& \ 0&1&1&1&1&0&0&0 \ -&-&-&-&-&-&-&- \ 1&1&1&0&0&0&0&1\
end{matrix}
$$

$$1111_2times 1111_2=(2^4-1)cdot (2^4-1)=2^8-2cdot 2^4+1\=2^8-2^5+1=2^5(2^3-1)+1
=2^5cdot111_2+1_2=11100000+1=11100001
$$

I used google sheets to organize the manual work, allowing for plenty of space on the left and top. If you need more workspace once you start you can insert rows and columns.

So the 'carry stuff' is above the red line and has to be included when you add the numbers between the two black lines. Each 'carry' goes in the next higher row.

Here is the work:

This is similar to the answer provided by астон вілла олоф мэллбэрг, but is more mechanized, stressing the organization of the work.

As you appear to have done, you can start the calculation by splitting up one of the numbers into its constituent powers of $2$ and then distributing the multiplication:

$$begin{aligned}color{blue}{1111_2}times1111_2&=(color{blue}{1000_2}+color{blue}{100_2}+color{blue}{10_2}+color{blue}{1_2})times1111_2\
&=(1000_2times1111_2)+(100_2times1111_2)+(10_2times1111_2)+(1_2times1111_2)\
&=1111000_2+111100_2+11110_2+1111_2quad(*)\
end{aligned}$$

In your third column from the right, where you get stuck, you have:

$$begin{aligned}underbrace{1}_{text{carried}}+1+1+1&=100_2
end{aligned}$$

For $100$, your bit is $0$ and you can carry the rest ($10_2$). Then, in the next column, you'd have:

$$begin{aligned}underbrace{10_2}_{text{carried}}+1+1+1+1&=110_2
end{aligned}$$

And just like before, your bit is $0$ but you carry the rest ($11_2$).

If you keep on with that method, you'd get:

$$begin{aligned}1111_2times1111_2&=color{red}{11100001_2}\
15times15,color{green}{✔}&=225,color{green}{✔}end{aligned}$$

(*) When multiplying any binary number by a power of $2$, you can just append the $0$'s of the power of $2$ onto the end of the other number. For example, $$begin{aligned}underbrace{(100_2)}_{text{power of 2}}times underbrace{(color{blue}{100110_2})}_{text{arbitrary binary no.}}
&=2^ntimes (color{blue}{2^{a}}+color{blue}{2^b}+color{blue}{2^c})\&=2^{a+n}+2^{b+n}+2^{c+n}\
&=10000000_2+10000_2+1000_2\&=10011000_2\
8times38,color{green}{✔}&=152,color{green}{✔}
end{aligned}$$