Generally, vector spaces "come" with a pre-specified field $\mathbf{K}$ (via the scalar multiplication function, $\cdot\colon \mathbf{K}\times\mathbf{V}\to\mathbf{V}$), and when discussing "subspaces" we only refer to subspaces that are themselves vector spaces over that field.
About the only way to sort of make sense of what you write is the following observation: if $F$ and $K$ are fields, $F\subseteq K$ (for example, $F=\mathbb{R}$ and $K=\mathbb{C}$; or $F=\mathbb{Q}$ and $K=\mathbb{R}$), and $\mathbf{V}$ is a vector space over $K$, then by restricting the scalar multiplication to only allow scalars from $F$ you also get a vector space; formally, it is a different vector space from the original (because the scalar multiplication function is different), though it is of course closely related. Any $\mathbf{W}$ that was a subspace of $\mathbf{V}$ when considered as a $\mathbf{K}$-vector space will also be a subspace of $\mathbf{V}$ when considering both as $\mathbf{F}$-vector spaces, but in general you may have subsets that are subspaces as $\mathbf{F}$-vector spaces but not as $\mathbb{C}$-vector spaces. For example, $\mathbf{C}$ is a vector space over $\mathbf{C}$; the only subspaces are $\mathbb{C}$ and $\{\mathbf{0}\}$. However, you can also consider $\mathbf{C}$ as an $\mathbf{R}$-vector space (allowing only real "scalars"). It's a different vector space (as a complex vector space, $\mathbb{C}$ has dimension $1$; as a real vector space, it has dimension $2$); both subspaces-as-complex-vector-space are still subspaces-as-real-vector space (if it was closed under multiplication by any complex number, it is also closed under multiplication by any real number; and vector addition is unchanged). But now we have lots of subspaces-as-real-vector-space that are not subspaces-as-complex-vector-space. For example, $\mathbf{W}=\{a+ai\mid a\in\mathbb{R}\}$ is easily seen to be a subspace if you only allow reals as scalars, but is not a subspace when working over the complex numbers.
When we do that, however, we are really talking about two different vector spaces: $\mathbf{V}$-as-a-$\mathbf{K}$-vector-space, and $\mathbf{V}$-as-an-$\mathbf{F}$-vector-space. We refer to the second as "the vector space obtained by restriction of scalars" (we "close the door" on scalars that are in $\mathbf{K}$ but not in $\mathbf{F}$, and don't consider them at all; we restrict what may be considered as a scalar). Though the two are intimitaly connected, we don't mix the structures: we need to talk separately about dimensions-over-$\mathbf{F}$ and dimensions-over-$\mathbf{K}$, we need to talk separately about $\mathbf{F}$-linear-maps and $\mathbf{K}$-linear-maps, etc. There is a one-way street, as anything that is linear-relative-to-$\mathbf{K}$ is also linear-relative-to-$\mathbf{F}$, but the road doesn't go the other way.
(Of course, the same set may be a vector space over many different fields in many different ways that have no relation to one another; the same set may "work" as a 1-dimensional vector space over $\mathbb{Q}$, as a 10-dimensional vector space over $\mathbb{Q}$, or as an infinite-dimensional vector space over a field with $49$ elements, but these structures will generally have absolutely nothing to do with one another. It's really best to think of the vector space as consisting of four things: a field $\mathbf{K}$ (often understood from context, but not always), a set $\mathbf{V}$, a function $+\colon\mathbf{V}\times\mathbf{V}\to\mathbf{V}$, and a function $\cdot\colon \mathbf{K}\times\mathbf{V}\to\mathbf{V}$, that satisfy certain properties (the axioms). )
In that sense, your question is not even well-posed: when you discuss "vector spaces over different fields", are you assuming the vector addition is the same? Are you assuming any connection between the fields and/or the scalar multiplications? And when you talk about "subspace", is it subspace relative to one field (which) or both? Dimension relative to what field? I suspect that trying to make the question sensible will clarify what your confusion might be, and help you realize that what you were perhaps fuzzily envisioning cannot occur at all.
You say "I've gathered that an isomorphism should preserve all of the structure, but I have a difficult time understanding how this comes from one-to-one and onto mapping."
In algebra, an isomorphism is defined to be a bijective homomorphism. It is the homomorphic quality which is structure preserving. (A bijection being a 1-1, onto mapping.)
Further, you state "I want to know if for any two vector spaces with an isomorphism between them, if there exists an isomorphism between each of their subspaces."
By definition, an isomorphism between two vector spaces will also act as an isomorphism between subspaces.
Best Answer
For finite dimensional vector spaces $V$ there is a natural bijection between subspaces of dimension $m$ of $V$ and subspaces of dimension $n-m$ of the dual of $V$. Given a subspace $A$ of dimension $m$, the corresponding subspace is the set of all linear functionals that vanish on $A$.
Edit: to incorporate Slade's comment and finish the story, if we have a nondegenerate bilinear form on $V$, not necessarily an inner product because this works over a completely general field, this gives us an isomorphism with $V^{\ast}$ and hence the bijection we seek.