Is it correct to say, "Time slows down the deeper we go in a gravitational field, because some of it is converted to spatial velocity"? Is it correct to say, "Time slows down the deeper we go in a gravitational field, because some of it is converted to spatial velocity"?
If we imagine a space-dimension on $x$-axis and Time on $y$-axis, then "time velocity" bends a little due to spacetime curvatures and hence only a component of it acts in the "time" direction and some of it is converted to spatial velocity. Is this line of thought correct, or am I messing up somewhere?
 A: I don't know why people downvote the question. It's obvious that you're not familiar enough with general relativity, but I understand your question.
In GR, the most important object is the metric, which tells us how to measure distances using some coordinates. Squared length of an infinitesimal line segment in spacetime is given by
$$ds^2 = g_{\mu \nu} dx^\mu dx^\nu =\left[ \matrix{ dx_0 &  dx_1 & dx_2 & dx_3} \right]\left[ \matrix{ g_{00} &  g_{01} & g_{02} & g_{03 } \\ g_{10} &  g_{11} & g_{12} & g_{13 } \\ g_{20} &  g_{21} & g_{22} & g_{23 } \\ g_{30} &  g_{31} & g_{32} & g_{33}} \right] \left[ \matrix{ dx_0 \\  dx_1 \\ dx_2 \\ dx_3} \right]$$
Here, $g_{\mu \nu}$ is called the metric tensor and it's a matrix that contains all the information we need in order to measure temporal and spatial distances.
Velocity in GR is described with a 4-vector known simply as the 4-velocity. It is given by
 $$U^\mu = \frac{dx^\mu}{d\tau} $$
$\tau$ stands for proper time, which is defined by $d\tau^2 = -ds^2$, but let's not worry about its meaning for now. Setting $c=1$ for simplicity, we can write the components of the vector explicitly:
$$U^\mu = (\gamma, \gamma \mathbf{u}) = \left[ \matrix{ \gamma \\ \gamma \frac{dx_1}{dt} \\ \gamma \frac{dx_2}{dt} \\ \gamma \frac{dx_3}{dt} } \right] $$
Here, $\gamma$ is the Lorentz factor given by $$\gamma = \frac{d\tau}{dt}$$
and it describes the ratio of change of proper time with respect to coordinate time. In other words, it describes how much you move in spacetime when you're not moving through space in that coordinate system. Or, if you will, it describes the velocity through time (time dilation, to be precise). 
The other three components describe velocity through space, multiplied by $\gamma$.
Following so far? If you're not, go search the unknown terms, you should find them easily!
Ok, with that out of the way, let's look what happens in the gravitational field near an uncharged non-rotating spherical object of mass $M$. The solution for this problem is most easily expressed using the Schwarzschild metric, given by
$$ds^2 =  - \left( 1 - \frac{2GM}{r} \right) dt^2 + \left( 1 - \frac{2GM}{r} \right)^{-1} dr^2 + r^2 d\Omega^2$$ 
Here, $G$ is Newton's gravitational constant and the coordinates we're using are known as Schwarzschild coordinates. $t$ is the time coordinate measured by an observer infinitely far away from the object and $r$ is the radial coordinate, measuring the circumference, divided by $2\pi$, of a sphere centered around the object. 
For simplicity, you can view $t$ as time and $r$ as the distance from the center of the object, but keep in mind that these coordinates have concrete definitions. $\Omega$ stands for angular coordinates, but let's not worry about them here.
By plugging it into the formula from the beginning, you can write the spacetime distance element (ignoring the angular part for simplicity, i.e. $d\Omega = 0$) using this matrix equation:
$$ds^2 = \left[ \matrix{ dt & dr } \right] \left[ \matrix{- \left( 1 - \frac{2GM}{r} \right) & 0 \\ 0 &\left( 1 - \frac{2GM}{r} \right)^{-1}} \right]  \left[ \matrix{ dt \\ dr } \right] $$
The velocity (again, ignoring the angular part) is then
$$   \left[ \matrix{ \frac{dt}{d \tau} \\ \frac{dr}{d \tau} } \right] = \left[ \matrix{ \frac{1}{\sqrt{  1 - \frac{2GM}{r}} }\\ \sqrt{ 1 - \frac{2GM}{r} } } \right] $$ 
So, when you move closer to the object, $r$ decreases and your velocity through time decreases while velocity through space increases. But don't consider distances $r\leq 2GM$ because you will encounter some trouble that is beyond the scope of this answer. (For completeness, let me just mention that objects with radius smaller than $2GM$ are black holes and their event horizon would be at $r_s=2GM$, known as the Schwarzschild radius)
So, the answer to your question is yes, but one has to be careful with words and be specific with their meaning.
I might have missed a minus sign somewhere, but it shouldn't affect the conclusion. I used the $(-,+,+,+)$ convention. Feel free to correct me or point to an error.
