I think that you’ve some misconceptions about both the workings of the algorithm and the reason it works.
Let’s look in detail at $37\cdot 15=555$. Here’s the correct table, in the arrangement that you used in your question, but with a little more detail. (Ignore the underlines and the $\text{Row}$ column for now.)
\begin{array}{c|cc}
\text{Row}&\text{Half}&&\text{Double}&\text{Remainder}\\ \hline
0&37&\times&\underline{15}&1\\
1&18&\times&30&0\\
2&9&\times&\underline{60}&1\\
3&4&\times&120&0\\
4&2&\times&240&0\\
5&1&\times&\underline{480}&1
\end{array}
There’s a remainder in the last line because $1$ would leave a remainder if you went on to halve it.
Ignore the $\text{Double}$ column for now. The first and last columns tell you that
$$\begin{align*}
37&=2\cdot 18+1\\
&=2(2\cdot 9+0)+1\\
&=2^2\cdot9+1\\
&=2^2(2\cdot4+1)+1\\
&=2^3\cdot4+2^2+1\\
&=2^3(2\cdot2+0)+2^2+1\\
&=2^4\cdot2+2^2+1\\
&=2^5+2^2+1\;.
\end{align*}\tag{1}$$
In other words, they show how to express $37$ as a sum of powers of $2$, i.e., how to write it in binary (base two) notation: $37=1\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+0\cdot2^1+1\cdot2^0$, so in binary it’s $100101$.
Now read the $\text{Remainder}$ column from bottom to top: $100101$. It’s exactly the same. And if you examine closely the calculations in $(1)$ and think about how they’re related to the original table, you’ll see that this will always work: the $\text{Remainder}$ column, read from bottom to top, gives you the binary representation of the number that you’re halving.
Now, finally, we’ll look at what’s happening in the $\text{Double}$ column. We want $37\cdot15$. We now know that $37=2^5+2^2+1$, so
$$\begin{align*}
37\cdot 15&=(2^5+2^2+1)\cdot 15\\
&=2^5\cdot15+2^2\cdot15+1\cdot15\;.
\end{align*}$$
Now $2^5\cdot15=\underbrace{2\cdot2\cdot2\cdot2\cdot2}_{5\text{ twos}}\cdot15$ is what you get when you double $15$ five times, $2^2\cdot15$ is what you get when you double $15$ twice, and of course $1\cdot15$ is just the original $15$. The $\text{Row}$ column of the table shows how many doublings (and halvings) have been performed to get to a given row, so are the numbers that I underlined in the $\text{Double}$ column. Thus,
$$\begin{align*}
37\cdot 15&=(2^5+2^2+1)\cdot 15\\
&=2^5\cdot15+2^2\cdot15+1\cdot15\\
&=480+60+15\\
&=555\;.
\end{align*}$$
Notice that these are also the numbers in the $\text{Double}$ column adjacent to $1$’s in the $\text{Remainder}$ column. That’s no accident: those $1$’s showed which powers of $2$ were needed to make up the multiplier $37$, and therefore which compound doublings of $15$ must be added to get the product.
The whole idea is use the repeated halving of the multiplier to write it as a sum of powers of two, because multiplying another number, like $15$, by a power of $2$ is easy: $2^na$ is what you get when you double $a$ $n$ times, and doubling is easy. The remainders of $1$ aren’t really lost: they tell you which powers of $2$ are actually needed in the binary representation of the multiplier and thereby tell you which doublings of the other number must be added together to get the product.