A vector space is an abelian group with some extra structure?
As you will recall, a (real) vector space is a set $A$ together with operations ${+}: A\times A\to A$ and ${\cdot}:\mathbb R\times A\to A$ that satisfy certain conditions. It happens that those conditions imply that $(A,{+})$ is an abelian group. So if you take any vector space and forget how scalar multiplication works, what is left is an abelian group.
Conversely, you can also imagine that the vector space arose by taking an abelian group and then defining a scalar multiplication for it -- this is the sense in which one can say that a vector space "is an abelian group with additional structure"; the "additional structure" is the scalar multiplication. Note that it is not all abelian groups that can be made into vector spaces in this way, and some abelian groups have more than one possible scalar multiplication -- the same group can be made into a several different vector spaces by choosing different multiplication operations.
When the exercise says "the group $V_1\times V_2$", it is to underscore that the $\times$ there is a product of groups, so when you see $V_1\times V_2$ you only have a single operation, namely the group operation (which will become the addition in the vector space you're going to construct).
When it asks you to prove that $V_1\times V_2$ "is a vector space", it's using sloppy language. What it really means is to prove that the product group can be made into a vector space by finding an appropriate scalar multiplication $\mathbb R\times(V_1\times V_2)\to V_1\times V_2$ and showing that it satisfies the conditions for a vector space. It is up to you to puzzle out a scalar multiplication that will make this work; it doesn't already exist as a part of the group product $V_1\times V_2$.
"A vector space is an abelian group with some extra structure." This follows from the axioms of a vector space. There are eight of them and the first four ones say exactly that $V$ is an abelian group:
- $\forall u,v,w\in V:\ u+(v+w)=(u+v)+w$,
- $\forall u\in V:\ u+0=u$,
- $\forall u\in V\ \exists -u\in V:\ u+(-u)=0$,
- $\forall u,v\in V:\ u+v=v+u$
1-3 say that $V$ is a group, 4 says it is abelian. Now what about the rest of axioms? They concern the multiplication by scalars and they are as follows:
- $\forall a\in\mathbb{K}\ \forall u,v\in V:\ a\cdot(u+v)=(a\cdot u)+(a\cdot v)$,
- $\forall a,b\in\mathbb{K}\ \forall u\in V:\ (a+b)\cdot u=(a\cdot u)+(b\cdot u)$,
- $\forall a,b\in\mathbb{K}\ \forall u\in V:\ a\cdot(b\cdot u)=(ab)\cdot u$,
- $\forall u\in V:\ 1_\mathbb{K}\cdot u=u$.
($1_\mathbb{K}$ is the multiplicative neutral element of the field $\mathbb{K}$). What do those axioms say? Let $End(V)$ be the ring of all endomorphisms of $V$ (ie. all group homomorphisms of $V$ into $V$). The multiplication in $End(V)$ is just a composition of functions and its neutral element is the identity function $id_V$: $id_V(u)=u$. Now what does it have to do with Axioms 5-8? They define a homomorphism from $\mathbb{K}$ into $End(V)$! How? Let $\varphi:\mathbb{K}\to End(V)$ be defined as follows $\varphi(a)(u)=a\cdot u$ where $a\in\mathbb{K}$ and $u\in V$. For each $a\in\mathbb{K}$ $\varphi(a)$ is a function from $V$ into $V$ -- it takes an element of $V$ and multiplies it by the scalar $a$. Axiom 5 say that for every $a\in\mathbb{K}$ $\varphi(a)$ is an endomorphism of $V$ ($\varphi(a)\in End(V)$), ie. $\varphi(a)$ is linear. Axioms 6-8 say that $\varphi$ is a homomorphisms between rings $\mathbb{K}$ (every field is a ring with multiplicative identity) and $End(V)$. This homomorphism $\varphi$ is the additional structure you are asking about.