The index of summation is ranging as
$$
a \le k \le n
$$
Replace it with its even and odd components
$$
k = 2j - i\quad \left| {\,i = 0,1} \right.
$$
Then for the even component it shall be
$$
\eqalign{
& i = 0\quad \Rightarrow \quad a \le 2j \le n\quad \Rightarrow \quad \left\lceil {{a \over 2}} \right\rceil \le j \le \left\lfloor {{n \over 2}} \right\rfloor
\quad \Rightarrow \cr
& \Rightarrow \quad \left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor \cr}
$$
and for the odd
$$
\eqalign{
& i = 1\quad \Rightarrow \quad a \le 2j - 1 \le n\quad \Rightarrow \quad \left\lceil {{{a + 1} \over 2}} \right\rceil \le j \le
\left\lfloor {{{n + 1} \over 2}} \right\rfloor \quad \Rightarrow \cr
& \Rightarrow \quad \left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor \cr}
$$
Note that for the lower bound we shall employ the ceiling and the floor for the upper. Then we can convert the ceiling to floor.
The sum over $k$ of $f(k)$ can be split into the sum over the even and the odd component
$$
\eqalign{
& \sum\limits_{a\, \le \,k\, \le \;n} {f(k)} = \sum\limits_{a\, \le \,k\, \le \;n} {\left. {f(k)\,} \right|_{\,k = 2j} + \left. {f(k)\,} \right|_{\,k = 2j - 1} } = \cr
& = \sum\limits_{a\, \le \,2j\, \le \;n} {f(2j)} + \sum\limits_{a\, \le \,2j - 1\, \le \;n} {f(2j - 1)} = \cr
& = \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {f(2j)}
+ \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {f(2j - 1)} \cr}
$$
and same for that of $(-1)^k f(k)$
$$
\eqalign{
& \sum\limits_{a\, \le \,k\, \le \;n} {\left( { - 1} \right)^{\,k} f(k)}
= \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {\left( { - 1} \right)^{\,2j} f(2j)}
+ \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {\left( { - 1} \right)^{\,2j - 1} f(2j - 1)} = \cr
& = \sum\limits_{\left\lfloor {{{a + 1} \over 2}} \right\rfloor \le j \le \left\lfloor {{n \over 2}} \right\rfloor } {f(2j)}
- \sum\limits_{\left\lfloor {{{a + 2} \over 2}} \right\rfloor \le j \le \left\lfloor {{{n + 1} \over 2}} \right\rfloor } {f(2j - 1)} \cr}
$$
Now, just subtract the two above to confirm the thesis.
--- addendum ----
Concerning your request on how to transform ceiling <-> floor, consider that
$$
\eqalign{
& n = 2\left\lceil {{n \over 2}} \right\rceil + 2\left\{ {{n \over 2}} \right\} = 2q + i\quad \left| \matrix{
\;n,q \in Z \hfill \cr
\,i = 0,1 \hfill \cr} \right. \cr
& \left\lceil {{n \over 2}} \right\rceil = \left\lceil {{{2q + i} \over 2}} \right\rceil
= \left\lceil {q + {i \over 2}} \right\rceil = q + \left\lceil {{i \over 2}} \right\rceil = q + i \cr
& \left\lfloor {{{n + 1} \over 2}} \right\rfloor = \left\lfloor {{{2q + 1 + i} \over 2}} \right\rfloor
= \left\lfloor {q + {{1 + i} \over 2}} \right\rfloor = q + \left\lfloor {{{1 + i} \over 2}} \right\rfloor = q + i \cr}
$$
and, in general
$$
\left\lceil {{n \over m}} \right\rceil = \left\lfloor {{{n + m - 1} \over m}} \right\rfloor \quad \left| \matrix{
\;n,m \in Z \hfill \cr
\;1 \le m \hfill \cr} \right.
$$
re. for details to this article.
Lucas' theorem does the job. Write $n=\sum\limits_0^q p^kn_k$ where $n_k$'s are the digits in the base $p$-representation of $n$. Since $n\gt p$, we must have $q\geq 1$. Then, by Lucas' theorem,
$$\binom np\equiv\binom{n_1}1\binom{n_0}0\equiv n_1\pmod p$$
Also, $$\lfloor n/p\rfloor=\lfloor(\sum_0^q p^kn_k)/p\rfloor=\lfloor n_0/p\rfloor+n_1+p(\cdots)=0+n_1+p(\cdots)\equiv n_1\pmod p$$
Thus, $$\binom np\equiv\left\lfloor\frac np\right\rfloor\pmod p$$
I'm not sure if an analogous argument would work for modulo prime powers of $p$ because Lucas' theorem modulo prime powers is a bit different from the usual statement, but here's a (sort of) generalization:
$$\binom n{p^k}\equiv\left\lfloor\frac n{p^k}\right\rfloor\equiv n_k\pmod p$$
where $k$ goes from $1$ to $q$ (also the trivial case of $k=0$ holds)
Best Answer
If $n$ is an integer and $t$ any real number, then it is straightforward to show that$\lfloor t + n \rfloor = \lfloor t \rfloor + n.$ Therefore, since $\lfloor x \rfloor$ is an integer, \begin{align*} \quad &\lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor \\ = &\lfloor y + x \rfloor - \lfloor x \rfloor + \lfloor x \rfloor \\ = &\lfloor x + y \rfloor. \end{align*}