Space Complexity measures the amount of space it takes to carry out a specific task (as a function of the input size). It is very similar to time-complexity, that measures the amount of time it would take to complete the task. In both cases, the complexity highly depends on the specific model of machine (RAM machine is faster than a double-tape Turing Machine, which is in turn faster than a single-tape Turing Machine).
Let's now focus on a specific simple model: Single-tape TM.
Example: Sorting
Consider a simple task: say, sorting. The input is a list of numbers $x=\{a1, a2, ...\}$ and the machine should write them in order. The size of the input is $|x|=n$.
One way to perform this task on a single-tape Turing machine, is to perform a Merge Sort: we look at the first number $a1$, split the list into two list, one for numbers that are larger than $a1$ and the second list with the other numbers (and then we repeat it recursively).
But how do we implement "spitting the list into two" on a single-tape TM?
A possible solution will mark the end of the input, then go over the list $n$ times, and every time it finds a number that should be on the second list, it will move it to the end of tape.
So, if we started with $n$ numbers, now we have a list of $2n$ places (some of which will be empty, since we "moved" the numbers to somewhere else). If we continue with this naive approach, we will keep increasing the space in use more and more, and will get a very large space complexity (say $n^2$ if we stupidly double the space at each phase of the recursion).
But there is a more efficient way to do it: after creating two lists, we can "close the empty gaps" that occurred when we moved elements. Since we never add or remove elements, the same amount of space we started with is always big enough to keep the intermediate lists.
For this approach we might need to maintain some counters that will tell us where each list begins and ends. Using smart methods, this can be done very efficiently, so that we will never use more than $\approx 2n$ cells of the tape. That is, the space complexity is $O(n)$ in this case.
(see also In-place merge sort, which is not exactly what I was talking about, but might give you more intuition)
Now you can also see that Space and Time are related. If we perform the space-efficient variant, the TM needs to do some extra-work of "closing gaps", marking start and end points, keep track of counters, etc. This extra work takes time.
It is very common that there is a tradeoff between space and time:
if the algorithm is very fast, it takes a lot of memory, and if it is very space-efficient then it takes more time. For the task of sorting, see a table comparing the time and space complexity of various sorting algorithms. (Note that in that table they count the space complexity excluding the input).
Space complexity and its relation to Time complexity
So now we can define the space complexity of "sorting" as the minimal space it takes to sort. That is, take the most-efficient (memory-wise) sorting algorithm. How much space does it take to sort $n$ elements? This is the space complexity of sorting.
More generally, we can define classes of tasks that take the same space complexity. For instance $DSPACE(n)$ contains all the tasks that can be solved using $O(n)$ memory. Said differently, it contains all the tasks whose space-complexity is at most $O(n)$
There is a strict hierarchy here: If we know that a task needs $O(n)$ space, it clearly cannot be in $DSPACE(1)$. More accurately, for any function $f(n)$,
$$ DSPACE( o(f(n)) \subsetneq DSPACE( f(n)).$$
Finally, we can relate time and space:
$$DTIME (f(n)) \subseteq DSPACE(f(n)),$$
since in $x$ time we can access at most $x$ cells of the tape. More interestingly, we know that
$$DSPACE(f(n)) \subseteq DTIME(2^{O(f(n))}).$$
That is, whatever task that we can solve using $x$ bits of memory, we can also solve using in at most $2^{O(x)}$ time-steps. This follows since a machine with $x$ memory has maximal number of $2^{O(x)}$ different configurations, so if it doesn't go into an infinite loop it must end in less than $2^{O(x)}$ time.
