Let $F_2$ be the equivalence relation defined on $2^{\mathbb{N}\times\mathbb{N}}$, where $fF_2g\Leftrightarrow_\text{def}\{f(m,-)|m\in\mathbb{N}\}=\{g(m,-)|m\in\mathbb{N}\}$. Define the subset $B_2=\{f\in 2^{\mathbb{N}\times\mathbb{N}}|n\neq m\Rightarrow f(n,-)\neq f(m,-)\}$. I aim to show that $F_2|_{B_2}$, the relation $F_2$ restricted to $B_2$, is Borel equivalent to $F_2$. Just in case:
For equivalence relations $E$ and $F$ I mean by $E\leq_BF$ a Borel reduction, i.e. $xEy\Leftrightarrow\theta(x)F\theta(y)$ for some Borel map $\theta$ and by Borel equivalent I mean "$E\leq_BF$ and $F\leq_BE$"
To show that $F_2|_{B_2}\leq_BF_2$ is not hard: Note that the inclusion $B_2\to2^{{\mathbb{N}\times\mathbb{N}}}$ is continuous, and hence also Borel, so $F_2|_{B_2}$ is in fact continuously reducible to $F_2$.
I find the other direction hard to prove, because there is a lot of information to keep track of. I have been given the hint to consider the operation $2^\mathbb{N}\times\mathbb{N}\ni(f,m)\mapsto((0^m1)^\frown f)$ where $((0^m1)^\frown f)$ denotes the string with zeros in the first $m$ entries, then a single 1 and then concatenated with $f$. I can tell that the hint is supposed to help me think of how to associate to any $f\in2^{\mathbb{N}\times\mathbb{N}}$, in a Borel way, a sequence $\widetilde{f}\in B_2$.
Here is what I have thought of: Given any $f\in2^{\mathbb{N}\times\mathbb{N}}$ consider the set $B(f)$ consisting of all $n$ such that $f(n,-)=f(m,-)$ for some $m\neq n$. In case $B(f)=\emptyset$ we already have $f\in B_2$ and we let $\widetilde{f}=f$. In case $B(f)$ is not empty, i.e. $f\not\in B_2$, we must systematically fix all $f(i,-)$ with $i\in B(f)$ so that these a distinct. For some reason I thought it be best to only fix $f(i,-)$ for $i\in B(f)$, which lead me to my first failed attempt: Let $i_0<i_1\dots$ be an enumeration $B(f)$. Exchange $f(i_\alpha,-)$ with $((0^\alpha1)^\frown f(i_\alpha,-))$. If I am not mistaken, this fixes all $f(i_\alpha,-)$, but in many cases this only causes new problems because $((0^\alpha1)^\frown f(i_\alpha,-))$ might very well equal $f(k,-)$ for some $k\not\in B(f)$. Then I thought that we might have to exchange $f(i,-)$ with $((0^i1)^\frown f(i,-))$ for all $i$ because then $\widetilde{f}\in B_2$ but it appears that $fF_2g\not\Rightarrow \widetilde{f}F_2|_{B_2}\widetilde{g}$, and I am not sure whether this is even Borel.
In other words I find it hard to figure out how to effectively assign to each $f\in 2^{\mathbb{N}\times\mathbb{N}}$ a $\widetilde{f}\in B_2$ in a Borel way so that $fF_2g\Leftrightarrow \widetilde{f}F_2|_{B_2}\widetilde{g}$. There are so many things to balance. I would appreciate hints, comments and suggestions on how to define $f\mapsto\widetilde{f}$ properly
Thank you in advance!
As you say, our starting point is to find a "natural" map $2^{\mathbb{N}\times\mathbb{N}}\rightarrow B_2$. Of course we'll then need to argue that this map is appropriate (i.e. is Borel and yields an appropriate reduction), but finding a good candidate map needs to happen first.
I think it will help to do the following. First, instead of $2^{\mathbb{N}\times\mathbb{N}}$ I'm going to talk in terms of $(2^\mathbb{N})^\mathbb{N}$, that is, sequences of reals instead of binary arrays. I think this makes things more intuitive. Second, and more substantively, I'll start by treating the simpler example of $F_2\vert_{B_1}$ where $B_1=\{f\in (2^\mathbb{N})^\mathbb{N}: ran(f)\mbox{ is infinite}\}$, that is, the sequences which take on infinitely many distinct values. $B_2$ meanwhile is the set of sequences which are injective. Note that $B_1$ is a Borel subset of $(2^\mathbb{N})^\mathbb{N}$ (this is a good exercise), so it's not too wild.
There's a natural way to "injectify" an element of $B_1$: simply "erase each term" which is equal to some earlier term, and then "collapse" the resulting sequence to get a genuine sequence of reals. For example, thinking about sequences of naturals instead of reals, this process would work as follows: given the sequence $$\alpha=3,1,6,2,1, 43,6,3,9,...$$ we'd first mark the terms which copy some earlier term $$3,1,6,2,\color{red}{1}, 42, \color{red}{6}, \color{red}{3},9, ...,$$ then get rid of them $$3,1,6,2,\Box,42,\Box,\Box,9,...,$$ and then collapse the result to get a genuine sequence $$\alpha'=3,1,6,2,42,9,....$$
Let $I:B_1\rightarrow B_2$ be the analogous injectification process for sequences of reals (with infinite range). We clearly have $$fF_2g\iff I(f)F_2I(g)$$ for $f,g\in B_1$, so if we can show that $I$ is Borel we'll have the desired $F_2\vert_{B_1}\le_BF_2\vert_{B_2}$. And at this point it helps to shift from thinking about $I$ as a function to $I$ as a set: "$I$ is Borel" is equivalent to "$\{(f,g)\in (2^\mathbb{N})^\mathbb{N}\times (2^\mathbb{N})^\mathbb{N}: g=I(f)\}$ is Borel."
So how do we say that $g=I(f)$? Well, we need to say each of the following:
$g$ is injective;
$ran(g)=ran(f)$; and
"elements of $ran(g)$ are enumerated by $g$ in the right order."
The first two conditions are easy to express appropriately: they correspond to "$\forall m_0<m_1\exists n(g(m_0)(n)\not=g(m_1)(n))$" and "$[\forall m\exists n\forall k(f(m)(k)=g(n)(k))]\wedge [\forall m\exists n\forall k(g(m)(k)=f(n)(k))]$" respectively. The third meanwhile is equivalent to "For each $m_0<m_1$ there is some $n$ such that $f(n)=g(m_0)$ and for each $j$, if $f(j)=g(m_1)$ then $j>n$." This also can be expressed using only quantifiers over natural numbers (I'll leave this as an exercise since it's tedious). So the whole expression "$g=I(f)$" is equivalent to a statement involving only quantification over natural numbers, and so the corresponding relation is Borel.
I'm using here a "definability-to-Borel-complexity" connection. This is an incredibly powerful tool for establishing Borel-ness; if you haven't seen it, it's absolutely worth your time. I'll add a citation for it in either Kechris or Moschovakis once I can find it.
OK, so we've proved (or outlined a proof of, anyways) the Borel equivalence of $F_2\vert_{B_1}$ and $F_2\vert_{B_2}$. It remains to replace the former with $F_2$ itself - the issue being how to handle those sequences with only finite range. For now I'll leave this as an exercise, although I can go into more detail if desired.