Your vector addition as I understand it is coordinate-wise addition whose output is a vector of length equal to the max length of the two inputs and whose entries are the sums of the corresponding entries with the caveat that if the entry is missing it is treated as zero.
Although addition can indeed be defined this way (and is done so in certain programming languages) it does not lend itself nicely to a group structure.
Recall that a group must have a "zero vector" $0$ such that $v+0=0+v=v$, that is to say adding by a zero vector will not change the object. Specifically note that $(0,0)$ is the proper "zero vector" here while $(0,0,0)$ is not since for example $(1,0)+(0,0,0)=(1,0,0)$ which is different (presumably) from $(1,0)$.
Recall that a group must have additive inverses for each element. That is, for any element $v$ you must have a $(-v)$ such that $v+(-v)=0$. But here, since any time you add with a 3-tuple you will get back another 3-tuple there is no way for us to have an additive inverse for a vector like $(1,0,0)$. At best we could have $(1,0,0)+(-1,0,0)=(0,0,0)$ but as mentioned in the previous paragraph this is not the zero vector in this context.
Since your set fails to have a group structure under the defined addition, your set does not qualify as a vector space.
As an alternative, if we were to amend our definition of what it means to be "equal" so as to allow $(1,0)=(1,0,0)$, letting two vectors be "equal" so long as every corresponding entry is either equal or if any missing entries exist that are missing in one but not the other that the corresponding entry is zero... then you will in fact have a vector space, but it is just the vectorspace commonly known as $\Bbb R^3$ and could have been written so as to contain only 3-tuples without having missed any elements.