The generic formula for the directional derivative of a function $f$ in the direction $u$ (for a unit vector) is $D_u f (x,y,z) = \nabla f(x,y,z) \cdot u$. For a vector, just do this to all the components.
Let's look at the example you give. Let's call $f_1(x,y,z) = x^3+y^2+z$, $\, f_2(x,y,z) = ze^x$, and $f_3(x,y,z) = xyz - 9xz$. Then the gradients are
$$ \begin {align*} \nabla f_1 &= (3x^2, \, 2y, \, 1) \\
\nabla f_2 &= (ze^x, \, 0, \, e^x) \\
\nabla f_3 &= (yz - 9z, \, xz, \, xy - 9x)
\end {align*}
$$
At your particular point $(1,2,3)$, these are:
$$ \begin {align*}
\nabla f_1(1,2,3) &= (3,4,1) \\
\nabla f_2(1,2,3) &= (3e,0,e) \\
\nabla f_3(1,2,3) &= (-27,3,-7)
\end {align*}
$$
The formula I mentioned above for directional derivative requires a UNIT vector. Since the vector you give, $u = (2,3,-5)$ is NOT a unit vector, we have to rescale it, and instead use the vector
$$ w = \frac{1}{\|u\|}u = \frac{1}{\sqrt{38}}(2,3,-5) $$
Now finally use the formula:
$$ \begin {align*}
D_w f_1(1,2,3) &= \frac{1}{\sqrt{38}}(2,3,-5) \cdot (3,4,1) = \frac{13}{\sqrt{38}} \\
D_w f_2(1,2,3) &= \frac{1}{\sqrt{38}}(2,3,-5) \cdot (3e,0,e) = \frac{e}{\sqrt{38}} \\
D_w f_3(1,2,3) &= \frac{1}{\sqrt{38}}(2,3,-5) \cdot (-27,3,-7) = \frac{98}{\sqrt{38}}
\end {align*}
$$
Putting it all together, the directional derivative of your vector function $v(x,y,z)$ in the direction u (also the same as direction w) is given by
$$ \frac{1}{\sqrt{38}} \left( \begin {array}{c} 13 \\ e \\ 98 \end{array} \right) $$
For the first question, write out the formula for projection of $\hat u$ onto $\vec t.$
Now write the formula for projection of $\vec t$ onto $\hat u.$
There is a dot product in both formulas, but other parts of the formulas are not the same as each other, are they? So even though both dot products give the same number, you get different results for each projection, as you expected.
For the second question, the component of $\vec t$ in the direction of $\hat u$ is the same as the component of $\vec t$ in the direction of $2\hat u$ or $0.1\hat u,$ is it not?
That is, if all you care about is the direction in which you want to take a component, the length of the vector that provides you with the direction shouldn't matter.
The way we make the length of the "direction pointing" vector not matter is
that whatever vector we are given, we replace it with a unit vector pointing in the same direction.
We need some kind of normalization like that because otherwise (as you should well know) the dot product $\vec u\cdot \vec t$ will be larger or smaller as the magnitude of $\vec u$ is larger or smaller.
More technically, the dot product satisfies the equation
$$ \vec u\cdot \vec t = \lVert\vec u\rVert\lVert\vec t\rVert \cos\theta,$$
where $\theta$ is the angle between the vectors.
Whereas (if $\theta$ is acute)
the projection of $\vec t$ onto $\vec u$ needs to be one leg of a right triangle with hypotenuse of length $\lVert\vec t\rVert$
and an angle $\theta$ between the projection vector and $\vec t.$
Draw the triangle; by the definition of cosine,
the length of the projection vector has to be $\lVert\vec t\rVert \cos\theta.$
This looks a lot like the right-hand side of the equation above, but the
factor $\lVert\vec u\rVert$ is missing.
So what to do about that pesky extra factor of $\lVert\vec u\rVert$
in the equation for $\vec u\cdot \vec t$?
Make sure it's always $1$ so it won't give you a wrong result.
That formula is my intuition about why in particular it makes sense to take a dot product with a unit vector.
Best Answer
Before we discuss vector-valued functions, it is important to interpret the dot-product between two fixed vectors. Most important, the dot product between two vectors is a scalar (typically a real or complex number). Geometrically, for vectors $u,v$ in Euclidean space, the dot product obeys the general formula $$ u \cdot v = \|u\| \|v\| \cos{\theta}$$ where $\theta$ is the angle between $u$ and $v$, and $\| \cdot \|$ indicates the length of the vector. For two vectors lying on a plane, it is a bit easier to visualize. Notice that if $\theta = \pi/2$, then the dot product is $0$, so orthogonal vectors yield a zero dot product. Furthermore, if $\theta = 0$, then then dot product is as large as possible if we hold the length of the vectors constant. What this means is that the size of the dot product combines two pieces of information:
If we just know the dot product, we can't separate these two; that is, if the dot product is big, it could be that the vectors are long, or that the vectors are very close to parallel, or both, whereas if the dot product is small, it could be that the vectors are short, or that they are very close to perpendicular, or both.
Now, turning to vector-valued functions, $r(t)$ and $u(t)$ each give you a vector at each particular "time" $t$, and so the function $r(t) \cdot u(t)$ is a scalar function that tells you the dot product at each given $t$; this product precisely measures the relationship between $r(t)$ and $u(t)$ as described above.