The answer on the site by @Qiaochu Yuan is terse and clear:
It means linear on the left:
$$\langle au + bv, \color{blue}w \rangle = a \langle u, \color{blue}w \rangle + b \langle v, \color{blue}w \rangle$$
and on the right:
$$\langle \color{blue}u, av + bw \rangle = a \langle \color{blue}u, v \rangle + b \langle \color{blue}u, w \rangle.$$
Now, I came across this definition at this point in this presentation on Lie groups by XylyXylyX:
$$\begin{align} \langle \color{blue}{v_1},\alpha v_2 + \beta v_3\rangle &= \alpha\langle \color{blue}{v_1},v_2\rangle + \beta\langle \color{blue}{v_1},v_3\rangle\\[2ex] \langle \alpha\color{blue}{v_1}+ \beta v_2, v_3 \rangle&= \langle \color{blue}{v_1},v_2\rangle \alpha+ \langle \color{blue}{v_1},v_3\rangle \beta\\[2ex] &=\langle \color{blue}{v_1},v_2\rangle \bar\alpha+ \langle \color{blue}{v_1},v_3\rangle \bar\beta \end{align}$$
attributed to Robert Gilmore Lie Groups, Lie Algebras, and Some of Their Applications. The presenter clearly explains that the definition is related to the treatment of quaternions as a field; however, I wonder if there a way to connect both definitions.
The answer by Yuan is certainly standard...for bilinear maps defined on vector spaces over the reals, and one of the key properties of the reals is that for any two real numbers, i.e., $a \cdot b = b \cdot a$, i.e., multiplication is commutative, as it is in all fields. That definition is the one used for bilinear functions on vector spaces over fields.
When you want to define something analogous for something like a vector space where the underlying set of scalars is not a field, but a "skew field" (one in which multiplication isn't necessarily commutative), there's a problem: when you "simplify" $\langle av, w \rangle$ by "pulling out the $a$", you have to decide whether to pull it out "in front" or "in the back", i.e., you have to decide whether $$ \langle av, w \rangle = a\langle v, w \rangle $$ or $$ \langle av, w \rangle = \langle v, w \rangle a. $$
If the underlying set of scalars has a commutative multiplication rule, these two are exactly the same, and the two definitions you've got are identical. If the multiplication is not commutative, they give different rules. Your author has chosen to use the second rule.
You might ask "Why is that the right rule to use when you're working with a the quaternions or some other skew-field?" and my guess is that when you use that definition, more of the standard theorems about bilinear forms in the "field" case carry over nicely.
This is a little related to inner products. You probably remember the inner product on $\Bbb R^n$ is defined by $$ \langle \pmatrix{a_1\\\vdots\\a_n},\pmatrix{b_1\\\vdots\\b_n} \rangle = a_1 b_1 + \ldots + a_n b_n. $$
But when you got to complex vector spaces, the inner product on $\Bbb C^n$ was defined by $$ \langle \pmatrix{a_1\\\vdots\\a_n},\pmatrix{b_1\\\vdots\\b_n} \rangle = a_1 \overline{b_1} + \ldots + a_n \overline{b_n}. $$ which is no longer bilinear, but is instead said to be "sesquilinear".
This had the advantage that $$ \langle x,x \rangle $$ then turned out to be a nonnegative real number, which you could reasonably call the "squared length" of $x$.
The two rules agree for $\Bbb R$, but differ when you consider $\Bbb C$, and the new notion (sesquilinear) used for $\Bbb C$ was designed to preserve useful ideas from the rule for $\Bbb R$.
Perhaps the rule you've given for quaternions should be called something like "skew-bilinear", but the author chose not to do that.
As a final note, you say that the quaternion reference says
\begin{align} \langle \alpha\color{blue}{v_1}+ \beta v_2, v_3 \rangle&= \langle \color{blue}{v_1},v_2\rangle \alpha+ \langle \color{blue}{v_1},v_3\rangle \beta\\[2ex] \end{align}
That's surely a typo, as it should be
\begin{align} \langle \alpha\color{blue}{v_1}+ \beta v_2, v_3 \rangle&= \langle \color{blue}{v_1},v_3\rangle \alpha+ \langle \color{blue}{v_2},v_3\rangle \beta \end{align}
If the first were correct, then picking $v_1 = 0$, $\alpha = \beta = 1$ we would conclude that for any vector $v_2$, we'd have \begin{align} \langle v_2, v_3 \rangle &= \langle 1 {\mathbf 0} + 1 v_2, v_3 \rangle \\ &= \langle { {\mathbf 0}},v_2\rangle 1+ \langle { {\mathbf 0}},v_3\rangle 1\\ &= 0, \end{align} which cannot be what's intended.