When, as in the second part of the question, DES's $f$ function is viewed as mapping $32$ input bits (a half block) to $32$ output bits (that get XORed with the other half block) for a fixed $48$-bit round key, $f$ is not injective, regardless of the fixed 48-bit round key that we consider(1). Since $f$ is a function from a finite set to itelf, it's thus not bijective, thus not surjective.
Therefore we assume from now on that the question asks why, and proof that, $f$ is surjective when viewed as mapping $32+48$ input bits (including round key) to $32$ output bits.
As of a proof: $f$ is defined in FIPS 46-3 as:
$$\begin{align}
f:\{0,1\}^{32}\times\{0,1\}^{48}&\to\ \{0,1\}^{32}\\
(R,K)\ &\mapsto f(R,K)\underset{\mathsf{def}}=P(S(E(R)\oplus K))\end{align}$$
where $E$ is the expansion, $S$ is the combination of S-boxes, and $P$ is the permutation. Fix to zero all $32$ bits of $R$, so that the input of the S-boxes is the $48$ bits of $K$. Further fix to zero the $16$ bits $1,6,7,12,\cdots,59,64$ of $K$, corresponding to the input bits of S-boxes that select an S-box row. This insures that the first row of each S-box is the one used. Each row of each S-box is a permutation. It follows that the function of the remaining $32$ bits of subkey $2,3,4,5,8,9,10,11,\cdots,60,61,62,63$ to the $32$ bits of output is a permutation. We can thus exhibit a $80$-bit input ($48$ bits of which zero) to any $32$-bit output of $f$, proving that $f$ is surjective(2).
As of the rationale (why): from a design perspective, the ideal for $f$ is arguably to behave like a random function of $80$ to $32$ bits. And with high likelihood, a random such function is surjective(3). The designers of DES did not want to chose one of the rare $f$ that are not surjective. In fact, they have selected an $f$ where the distribution of the outputs is exactly even, which is antagonist to not surjective.
(1) See my question and Thomas Pornin's un-challenged answer. In a nutshell his proof uses that each of the 4 rows of the 8 S-boxes is a permutation to show that the problem of checking surjectivity of $f$ for a given subkey is independent of $32$ out of $48$ bits of the subkey. He uses a program to find a collision for each of the $2^{16}$ remaining classes of subkeys, proving that $f$ is not injective.
(2) Further, we can prove that each $32$-bit output has exactly $2^{48}$ matching inputs. Sketch: For each $32$-bit output, for each $32$-bit $R$, each of $2^{16}$ choices of S-box rows matches a unique distinct choice of bits $1,6,7,12,\cdots,59,64$ of $K$, and then a single choice of the other $32$ bits of $K$.
(3) Non-rigorous argument: any fixed of the $2^{32}$ outputs is not reached by any of the $2^{80}$ inputs with probability $\approx 2^{-48}$, so under the approximation that these $2^{32}$ events are independent the probability that a random function is not surjective is $\approx 2^{-16}$.
