The basic idea is this. You have an input tape with $\#w_1\#w_2\#\ldots\#w_n$, where the $w_k$ are strings over $\{0,1\}$. You set a left marker, indicated in red, at the first $\#$:
$$\color{red}{\#}w_1\#w_2\#\ldots\#w_n$$
Then you scan to the right until you reach a $\#$ and set a right marker, indicated in blue:
$$\color{red}{\#}w_1\color{blue}{\#}w_2\#\ldots\#w_n$$
Now read the first character of $w_1$, mark it as read, and scan forward to compare it with the first character of $w_2$; if they’re equal, mark the first character of $w_2$ as read and scan back to find the second character of $w_1$. (You can find it, because you marked the first character as read.) Mark it read and scan forward to the second character of $w_2$; if they’re equal, mark the second character of $w_2$ read and scan back to the third character of $w_1$. Continue in this fashion until one of two things happens. If every character of $w_1$ matches the corresponding character of $w_2$, and $w_2$ has no extra characters, reject the string. Otherwise, go back and erase the reading marks that you made in $w_1$ and $w_2$ and move the right marker to the next $\#$:
$$\color{red}{\#}w_1\#w_2\color{blue}{\#}\ldots\#w_n$$
Repeat; this time you’ll be comparing $w_1$ with $w_3$. If you get a match, you’ll reject the input, and if not, you’ll advance the blue marker again. If $w_1$ is different from $w_2,\dots,w_n$, you’ll eventually reach a blank instead of a $\#$ when you try to advance the blue marker; at that point you go back and advance the red marker and re-initialize the blue marker to the next $\#$ to the right:
$$\#w_1\color{red}{\#}w_2\color{blue}{\#}w_3\#\ldots\#w_n$$
The next full cycle compares $w_2$ successively with $w_3,w_4,\dots,w_n$, rejecting the input if a match is found. If no match is found, you advance the red marker another step, reset the blue marker, and repeat.
I think that it is not a "real conceptual problem".
But if you want an unamboguous coding, you can see Wang coding, described in George Boolos & John Burgess & Richard Jeffrey, Computability and Logic (5th ed - 2007) : Ch.8.1 Coding Turing Computations, page 88-on.
Best Answer
Let $A$ and $B$ be languages. A string $w$ is in the concatenation of $A$ and $B$ if you can write it as $w= ab$, where $a\in A$ and $b\in B$. This means that you can split $w$ into two halves. The first half is a string from $A$ and the second half is a string from $B$.
If you have two decideable languages $A$ and $B$, you can make a decider for $A$ concatenated with $B$ as follows:
Given a two-tape nondeterministic Turing machine,
Everything else is just bookkeeping: